domain
listlengths 0
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float64 5
5
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stringlengths 48
1.74k
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stringlengths 3
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|---|---|---|---|---|---|
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5
|
In right triangle $A B C$, a point $D$ is on hypotenuse $A C$ such that $B D \perp A C$. Let $\omega$ be a circle with center $O$, passing through $C$ and $D$ and tangent to line $A B$ at a point other than $B$. Point $X$ is chosen on $B C$ such that $A X \perp B O$. If $A B=2$ and $B C=5$, then $B X$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100 a+b$.
|
Note that since $A D \cdot A C=A B^{2}$, we have the tangency point of $\omega$ and $A B$ is $B^{\prime}$, the reflection of $B$ across $A$. Let $Y$ be the second intersection of $\omega$ and $B C$. Note that by power of point, we have $B Y \cdot B C=B B^{\prime 2}=4 A B^{2} \Longrightarrow B Y=\frac{4 A B^{2}}{B C}$. Note that $A X$ is the radical axis of $\omega$ and the degenerate circle at $B$, so we have $X B^{2}=X Y \cdot X C$, so $$B X^{2}=(B C-B X)(B Y-B X)=B X^{2}-B X(B C+B Y)+B C \cdot B Y$$ This gives us $$B X=\frac{B C \cdot B Y}{B C+B Y}=\frac{4 A B^{2} \cdot B C}{4 A B^{2}+B C^{2}}=\frac{80}{41}$$
|
8041
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear?
|
There are $8!/(4!2!2!)=420$ ways to order the letters. If the permuted letters contain "HMMT", there are $5 \cdot 4!/ 2!=60$ ways to order the other letters, so we subtract these. However, we have subtracted "HMMTHMMT" twice, so we add it back once to obtain 361 possibilities.
|
361
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other",
"Mathematics -> Applied Mathematics -> Statistics -> Mathematical Statistics"
] | 5
|
In a game similar to three card monte, the dealer places three cards on the table: the queen of spades and two red cards. The cards are placed in a row, and the queen starts in the center; the card configuration is thus RQR. The dealer proceeds to move. With each move, the dealer randomly switches the center card with one of the two edge cards (so the configuration after the first move is either RRQ or QRR). What is the probability that, after 2004 moves, the center card is the queen?
|
If the probability that the queen is the center card after move n is $p_{n}$, then the probability that the queen is an edge card is $1-p_{n}$, and the probability that the queen is the center card after move $n+1$ is $p_{n+1}=\left(1-p_{n}\right) / 2$. This recursion allows us to calculate the first few values of $p_{n}$. We might then notice in $1,0, \frac{1}{2}, \frac{1}{4}, \frac{3}{8}, \frac{5}{16}, \frac{11}{32}, \cdots$, that the value of each fraction is close to $1 / 3$, and getting closer for larger $n$. In fact subtracting $1 / 3$ from each fraction yields $\frac{2}{3},-\frac{1}{3}, \frac{1}{6},-\frac{1}{12}, \frac{1}{24},-\frac{1}{48}, \cdots$. This suggests the formula $p_{n}=\frac{1}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)^{n}$, and one can then prove that this formula is in fact correct by induction. Thus, $p(2004)=\frac{1}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)^{2004}=\frac{1}{3}+\frac{1}{3 \cdot 2^{2003}}$. The recurrence can also be solved without guessing - by generating functions, for example, or by using the fundamental theorem of linear recurrences, which ensures that the solution is of the form $p_{n}=a+b(-1 / 2)^{n}$ for some constants $a, b$.
|
1/3 + 1/(3 \cdot 2^{2003})
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5
|
You have six blocks in a row, labeled 1 through 6, each with weight 1. Call two blocks $x \leq y$ connected when, for all $x \leq z \leq y$, block $z$ has not been removed. While there is still at least one block remaining, you choose a remaining block uniformly at random and remove it. The cost of this operation is the sum of the weights of the blocks that are connected to the block being removed, including itself. Compute the expected total cost of removing all the blocks.
|
Note that the total cost is the total number of ordered pairs $(x, y)$ with $1 \leq x, y \leq 6$ such that $x$ and $y$ are connected right before $x$ gets removed. The probability that blocks $x$ and $y$ are connected just before block $x$ is removed is simply $\frac{1}{|x-y|+1}$, since all of the $|x-y|+1$ relevant blocks are equally likely to be removed first. Summing over $1 \leq x, y \leq 6$, combining terms with the same value of $|x-y|$, we get $\frac{2}{6}+\frac{4}{5}+\frac{6}{4}+\frac{8}{3}+\frac{10}{2}+6=\frac{163}{10}$.
|
\frac{163}{10}
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5
|
How many orderings $(a_{1}, \ldots, a_{8})$ of $(1,2, \ldots, 8)$ exist such that $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}-a_{8}=0$ ?
|
We can divide the numbers up based on whether they have a + or - before them. Both the numbers following +'s and -'s must add up to 18. Without loss of generality, we can assume the + 's contain the number 1 (and add a factor of 2 at the end to account for this). The possible 4-element sets containing a 1 which add to 18 are $\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},\{1,4,6,7\}$. Additionally, there are 4! ways to order the numbers following a + and 4! ways to order the numbers following a -. Thus the total number of possibilities is $4 \times 2 \times 4!\times 4!=4608$.
|
4608
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
Compute \( \frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \frac{4^{3}-1}{4^{3}+1} \cdot \frac{5^{3}-1}{5^{3}+1} \cdot \frac{6^{3}-1}{6^{3}+1} \).
|
Use the factorizations \( n^{3}-1=(n-1)\left(n^{2}+n+1\right) \) and \( n^{3}+1=(n+1)\left(n^{2}-n+1\right) \) to write \( \frac{1 \cdot 7}{3 \cdot 3} \cdot \frac{2 \cdot 13}{4 \cdot 7} \cdot \frac{3 \cdot 21}{5 \cdot 13} \cdot \frac{4 \cdot 31}{6 \cdot 21} \cdot \frac{5 \cdot 43}{7 \cdot 31}=\frac{1 \cdot 2 \cdot 43}{3 \cdot 6 \cdot 7}=\frac{43}{63} \).
|
43/63
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Calculus -> Integral Calculus -> Techniques of Integration -> Single-variable"
] | 5
|
Evaluate the infinite sum $\sum_{n=0}^{\infty}\binom{2 n}{n} \frac{1}{5^{n}}$.
|
First Solution: Note that $\binom{2 n}{n} =\frac{(2 n)!}{n!\cdot n!}=\frac{(2 n)(2 n-2)(2 n-4) \cdots(2)}{n!} \cdot \frac{(2 n-1)(2 n-3)(2 n-5) \cdots(1)}{n!} =2^{n} \cdot \frac{(-2)^{n}}{n!}(-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}-2) \cdots(-\frac{1}{2}-n+1) =(-4)^{n}\binom{-\frac{1}{2}}{n}$. Then, by the binomial theorem, for any real $x$ with $|x|<\frac{1}{4}$, we have $(1-4 x)^{-1 / 2}=\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}(-4 x)^{n}=\sum_{n=0}^{\infty}\binom{2 n}{n} x^{n}$. Therefore, $\sum_{n=0}^{\infty}\binom{2 n}{n}(\frac{1}{5})^{n}=\frac{1}{\sqrt{1-\frac{4}{5}}}=\sqrt{5}$. Second Solution: Consider the generating function $f(x)=\sum_{n=0}^{\infty}\binom{2 n}{n} x^{n}$. It has formal integral given by $g(x)=I(f(x))=\sum_{n=0}^{\infty} \frac{1}{n+1}\binom{2 n}{n} x^{n+1}=\sum_{n=0}^{\infty} C_{n} x^{n+1}=x \sum_{n=0}^{\infty} C_{n} x^{n}$ where $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. Let $h(x)=\sum_{n=0}^{\infty} C_{n} x^{n}$; it suffices to compute this generating function. Note that $1+x h(x)^{2}=1+x \sum_{i, j \geq 0} C_{i} C_{j} x^{i+j}=1+x \sum_{k \geq 0}(\sum_{i=0}^{k} C_{i} C_{k-i}) x^{k}=1+\sum_{k \geq 0} C_{k+1} x^{k+1}=h(x)$ where we've used the recurrence relation for the Catalan numbers. We now solve for $h(x)$ with the quadratic equation to obtain $h(x)=\frac{1 / x \pm \sqrt{1 / x^{2}-4 / x}}{2}=\frac{1 \pm \sqrt{1-4 x}}{2 x}$. Note that we must choose the - sign in the $\pm$, since the + would lead to a leading term of $\frac{1}{x}$ for $h$ (by expanding $\sqrt{1-4 x}$ into a power series). Therefore, we see that $f(x)=D(g(x))=D(x h(x))=D(\frac{1-\sqrt{1-4 x}}{2})=\frac{1}{\sqrt{1-4 x}}$ and our answer is hence $f(1 / 5)=\sqrt{5}$.
|
\sqrt{5}
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Probability -> Other"
] | 5
|
Diana is playing a card game against a computer. She starts with a deck consisting of a single card labeled 0.9. Each turn, Diana draws a random card from her deck, while the computer generates a card with a random real number drawn uniformly from the interval $[0,1]$. If the number on Diana's card is larger, she keeps her current card and also adds the computer's card to her deck. Otherwise, the computer takes Diana's card. After $k$ turns, Diana's deck is empty. Compute the expected value of $k$.
|
By linearity of expectation, we can treat the number of turns each card contributes to the total independently. Let $f(x)$ be the expected number of turns a card of value $x$ contributes (we want $f(0.9)$). If we have a card of value $x$, we lose it after 1 turn with probability $1-x$. If we don't lose it after the first turn, which happens with probability $x$, then given this, the expected number of turns this card contributes is $f(x)+\frac{1}{x} \int_{0}^{x} f(t) d t$. Thus, we can write the equation $$f(x)=1+x f(x)+\int_{0}^{x} f(t) d t$$ Differentiating both sides gives us $$f^{\prime}(x)=x f^{\prime}(x)+f(x)+f(x) \Longrightarrow \frac{f^{\prime}(x)}{f(x)}=\frac{2}{1-x}$$ Integrating gives us $\ln f(x)=-2 \ln (1-x)+C \Longrightarrow f(x)=\frac{e^{C}}{(1-x)^{2}}$. Since $f(0)=1$, we know that $C=0$, so $f(x)=(1-x)^{-2}$. Thus, we have $f(0.9)=(1-0.9)^{-2}=100$.
|
100
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Number Theory -> Prime Numbers"
] | 5
|
Define $\phi^{!}(n)$ as the product of all positive integers less than or equal to $n$ and relatively prime to $n$. Compute the number of integers $2 \leq n \leq 50$ such that $n$ divides $\phi^{!}(n)+1$.
|
Note that, if $k$ is relatively prime to $n$, there exists a unique $0<k^{-1}<n$ such that $k k^{-1} \equiv 1(\bmod n)$. Hence, if $k^{2} \not \equiv 1(\bmod n)$, we can pair $k$ with its inverse to get a product of 1. If $k^{2} \equiv 1(\bmod n)$, then $(n-k)^{2} \equiv 1(\bmod n)$ as well, and $k(n-k) \equiv-k^{2} \equiv-1(\bmod n)$. Hence these $k$ can be paired up as well, giving products of -1. When $n \neq 2$, there is no $k$ such that $k^{2} \equiv 1$ $(\bmod n)$ and $k \equiv n-k(\bmod n)$, so the total product $(\bmod n)$ is $(-1)^{\frac{m}{2}}$, where $m$ is the number of $k$ such that $k^{2} \equiv 1(\bmod n)$. For prime $p$ and positive integer $i$, the number of solutions to $k^{2} \equiv 1\left(\bmod p^{i}\right)$ is 2 if $p$ is odd, 4 if $p=2$ and $i \geq 3$, and 2 if $p=i=2$. So, by Chinese remainder theorem, if we want the product to be -1, we need $n=p^{k}, 2 p^{k}$, or 4. We can also manually check the $n=2$ case to work. Counting the number of integers in the allowed range that are of one of these forms (or, easier, doing complementary counting), we get an answer of 30. (Note that this complicated argument basically reduces to wanting a primitive root.)
|
30
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Prime Numbers",
"Mathematics -> Number Theory -> Factorization"
] | 5
|
A positive integer $n$ is loose if it has six positive divisors and satisfies the property that any two positive divisors $a<b$ of $n$ satisfy $b \geq 2 a$. Compute the sum of all loose positive integers less than 100.
|
Note that the condition in the problem implies that for any divisor $d$ of $n$, if $d$ is odd then all other divisors of $n$ cannot lie in the interval $\left[\left\lceil\frac{d}{2}\right\rceil, 2 d-1\right]$. If $d$ is even, then all other divisors cannot lie in the interval $\left[\frac{d}{2}+1,2 d-1\right]$. We first find that $n$ must be of the form $p^{5}$ or $p^{2} q$ for primes $p$ and $q$. If $n=p^{5}$, the only solution is when $p=2$ and $n=32$. Otherwise, $n=p^{2} q$. Since $100>n>p^{2}$, so $p \leq 7$. Now we can do casework on $p$. When $p=2$, we find that $q$ cannot lie in $[2,3]$ or $[3,7]$, so we must have $q \geq 11$. All such values for $q$ work, giving solutions $n=44,52,68,76,92$. When $p=3$, we find that $q$ cannot lie in $[2,5]$ or $[5,17]$, so we must have that $q \geq 19$, so there are no solutions in this case. When $p=5$ or $p=7$, the only solution occurs when $q=2$ (since otherwise $n>100$). This gives us the solutions $n=50$ and $n=98$. Adding these values of $n$ gives 512.
|
512
|
HMMT_2
|
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 5
|
For a real number $x$, let $[x]$ be $x$ rounded to the nearest integer and $\langle x\rangle$ be $x$ rounded to the nearest tenth. Real numbers $a$ and $b$ satisfy $\langle a\rangle+[b]=98.6$ and $[a]+\langle b\rangle=99.3$. Compute the minimum possible value of $[10(a+b)]$.
|
Without loss of generality, let $a$ and $b$ have the same integer part or integer parts that differ by at most 1, as we can always repeatedly subtract 1 from the larger number and add 1 to the smaller to get another solution. Next, we note that the decimal part of $a$ must round to .6 and the decimal part of $b$ must round to .3. We note that $(a, b)=(49.55,49.25)$ is a solution and is clearly minimal in fractional parts, giving us $[10(a+b)]=988$.
|
988
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Factorization",
"Mathematics -> Algebra -> Prealgebra -> Factorials -> Other"
] | 5
|
Find the largest number $n$ such that $(2004!)!$ is divisible by $((n!)!)!$.
|
For positive integers $a, b$, we have $$a!\mid b!\quad \Leftrightarrow a!\leq b!\quad \Leftrightarrow \quad a \leq b$$ Thus, $$((n!)!)!\mid(2004!)!\Leftrightarrow(n!)!\leq 2004!\Leftrightarrow n!\leq 2004 \Leftrightarrow n \leq 6$$
|
6
|
HMMT_2
|
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Applied Mathematics -> Probability -> Other"
] | 5
|
In the Cartesian plane, let $A=(0,0), B=(200,100)$, and $C=(30,330)$. Compute the number of ordered pairs $(x, y)$ of integers so that $\left(x+\frac{1}{2}, y+\frac{1}{2}\right)$ is in the interior of triangle $A B C$.
|
We use Pick's Theorem, which states that in a lattice polygon with $I$ lattice points in its interior and $B$ lattice points on its boundary, the area is $I+B / 2-1$. Also, call a point center if it is of the form $\left(x+\frac{1}{2}, y+\frac{1}{2}\right)$ for integers $x$ and $y$. The key observation is the following - suppose we draw in the center points, rotate $45^{\circ}$ degrees about the origin and scale up by $\sqrt{2}$. Then, the area of the triangle goes to $2 K$, and the set of old lattice points and center points becomes a lattice. Hence, we can also apply Pick's theorem to this new lattice. Let the area of the original triangle be $K$, let $I_{1}$ and $B_{1}$ be the number of interior lattice points and boundary lattice points, respectively. Let $I_{c}$ and $B_{c}$ be the number of interior and boundary points that are center points in the original triangle. Finally, let $I_{2}$ and $B_{2}$ be the number of interior and boundary points that are either lattice points or center points in the new triangle. By Pick's Theorem on both lattices, $$\begin{aligned} K & =I_{1}+B_{1} / 2-1 \\ 2 K & =I_{2}+B_{2} / 2-1 \\ \Longrightarrow\left(I_{2}-I_{1}\right) & =K-\frac{B_{1}-B_{2}}{2} \\ \Longrightarrow I_{c} & =K-\frac{B_{c}}{2} \end{aligned}$$ One can compute that the area is 31500. The number of center points that lie on on $A B, B C$, and $C A$ are 0,10, and 30, respectively. Thus, the final answer is $31500-\frac{0+10+30}{2}=31480$.
|
31480
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
The function $f(x)$ is of the form $a x^{2}+b x+c$ for some integers $a, b$, and $c$. Given that $\{f(177883), f(348710), f(796921), f(858522)\} = \{1324754875645,1782225466694,1984194627862,4388794883485\}$ compute $a$.
|
We first match the outputs to the inputs. To start, we observe that since $a \geq 0$ (since the answer to the problem is nonnegative), we must either have $f(858522) \approx 4.39 \cdot 10^{12}$ or $f(177883) \approx 4.39 \cdot 10^{12}$. However, since 858522 is relatively close to 796921, the first case is unrealistic, meaning that the second case must be true. Now, looking $\bmod 2$, we find that $f(796921) \approx 1.32 \cdot 10^{12}$. Additionally, we find that $\bmod 5, f(1) \equiv f(3) \equiv 0(\bmod 5)$, so $f(x) \equiv a(x-1)(x-3)(\bmod 5)$. Modulo 5, we now have $\{3 a, 4 a\}=\{f(0), f(2)\}=\{2,4\}$, so it follows that $a \equiv 3(\bmod 5), f(349710) \approx 1.78 \cdot 10^{12}$ and $f(858522) \approx 1.98 \cdot 10^{12}$. There are several ways to finish from here. One (somewhat tedious) method is to use mod 9, which tells us that $f(7)=7, f(5)=8, f(3)=4$, which tells you that $a \equiv 5(\bmod 9)$ (take a finite difference). This tells you that $a \equiv 23(\bmod 45)$, and $a \geq 68$ can be ruled out for being too large. Another method is to work with the numbers themselves. One way to do this is to note that for quadratic polynomials, $$f^{\prime}\left(\frac{x+y}{2}\right)=\frac{f(y)-f(x)}{y-x}$$ Using this for $\{177883,348710\}$ and $\{796921,858522\}$, we find that $f^{\prime}(260000) \approx-1500000$ and $f^{\prime}(830000) \approx 1000000$. Thus $f^{\prime}$ (which we know must be linear with slope $2 a$) has slope just less than 50. Either way, we find that $a=23$. The actual polynomial is $8529708870514-27370172 x+23 x^{2}$.
|
23
|
HMMT_2
|
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 5
|
Regular polygons $I C A O, V E N T I$, and $A L B E D O$ lie on a plane. Given that $I N=1$, compute the number of possible values of $O N$.
|
First, place $A L B E D O$. We then note that $I C A O$ has two orientations, both of which have $I$ on $E O$. Next, we note that for any given orientation of $I C A O$, the two orientations of $V E N T I$ have $N$ symmetric to line $E I$. Thus, for any given orientation of $I C A O$, we have that $O N$ is the same in both orientations of $V E N T I$, which gives a total of 2 possible values for $O N$.
|
2
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
Suppose that there exist nonzero complex numbers $a, b, c$, and $d$ such that $k$ is a root of both the equations $a x^{3}+b x^{2}+c x+d=0$ and $b x^{3}+c x^{2}+d x+a=0$. Find all possible values of $k$ (including complex values).
|
Let $k$ be a root of both polynomials. Multiplying the first polynomial by $k$ and subtracting the second, we have $a k^{4}-a=0$, which means that $k$ is either $1,-1, i$, or $-i$. If $a=b=c=d=1$, then $-1, i$, and $-i$ are roots of both polynomials. If $a=b=c=1$ and $d=-3$, then 1 is a root of both polynomials. So $k$ can be $1,-1, i$, and $-i$.
|
1,-1, i,-i
|
HMMT_2
|
[
"Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5
|
Compute the number of positive real numbers $x$ that satisfy $\left(3 \cdot 2^{\left\lfloor\log _{2} x\right\rfloor}-x\right)^{16}=2022 x^{13}$.
|
Let $f(x)=3 \cdot 2^{\left\lfloor\log _{2} x\right\rfloor}-x$. Note that for each integer $i$, if $x \in\left[2^{i}, 2^{i+1}\right)$, then $f(x)=3 \cdot 2^{i}-x$. This is a line segment from $\left(2^{i}, 2^{i+1}\right)$ to $\left(2^{i+1}, 2^{i}\right)$, including the first endpoint but not the second. Now consider the function $f(x)^{16} / x^{13}$. This consists of segments of decreasing functions connecting $\left(2^{i}, 2^{3 i+16}\right)$ and $\left(2^{i+1}, 2^{3 i-13}\right)$. Note that for each $-1 \leq i \leq 7$, we have that $2^{3 i-13} \leq 2^{10}<2022<2^{11} \leq 2^{3 i+16}$. This gives us 9 solutions in total.
|
9
|
HMMT_2
|
[
"Mathematics -> Algebra -> Abstract Algebra -> Other"
] | 5
|
Suppose that $P(x, y, z)$ is a homogeneous degree 4 polynomial in three variables such that $P(a, b, c)=P(b, c, a)$ and $P(a, a, b)=0$ for all real $a, b$, and $c$. If $P(1,2,3)=1$, compute $P(2,4,8)$.
|
Since $P(a, a, b)=0,(x-y)$ is a factor of $P$, which means $(y-z)$ and $(z-x)$ are also factors by the symmetry of the polynomial. So, $$\frac{P(x, y, z)}{(x-y)(y-z)(z-x)}$$ is a symmetric homogeneous degree 1 polynomial, so it must be $k(x+y+z)$ for some real $k$. So, the answer is $$\frac{P(2,4,8)}{P(1,2,3)}=\frac{(2+4+8)(2-4)(4-8)(8-2)}{(1+2+3)(1-2)(2-3)(3-1)}=56$$
|
56
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the solutions to the equation $x^{4}+3 x^{3}+3 x^{2}+3 x+1=0$. Then $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|+\left|z_{4}\right|$ can be written as $\frac{a+b \sqrt{c}}{d}$, where $c$ is a square-free positive integer, and $a, b, d$ are positive integers with $\operatorname{gcd}(a, b, d)=1$. Compute $1000 a+100 b+10 c+d$.
|
Note that $x=0$ is clearly not a solution, so we can divide the equation by $x^{2}$ to get $\left(x^{2}+2+\frac{1}{x^{2}}\right)+3\left(x+\frac{1}{x}\right)+1=0$. Letting $y=x+\frac{1}{x}$, we get that $y^{2}+3 y+1=0$, so $y=x+\frac{1}{x}=\frac{-3 \pm \sqrt{5}}{2}$. Since $\frac{-3+\sqrt{5}}{2}$ has absolute value less than 2, the associated $x$ are on the unit circle, and thus the two solutions for $x$ in this case each have magnitude 1. For $\frac{-3-\sqrt{5}}{2}$, the roots are negative reals that are reciprocals of each other. Thus, the sum of their absolute values is the absolute value of their sum, which is $\frac{3+\sqrt{5}}{2}$. Thus, the sum of the magnitudes of the four solutions are $1+1+\frac{3+\sqrt{5}}{2}=\frac{7+\sqrt{5}}{2}$.
|
7152
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5
|
$\mathbf{7 3 8 , 8 2 6}$. This can be arrived at by stepping down, starting with finding how many combinations are there that begin with a letter other than V or W , and so forth. The answer is $\frac{8 \cdot 9!}{2 \cdot 2}+\frac{4 \cdot 7!}{2}+4 \cdot 6!+4 \cdot 4!+3!+2!+2!=738826$.
|
The number of combinations is 738826.
|
738826
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5
|
Find all real solutions to $x^{4}+(2-x)^{4}=34$.
|
Let $y=2-x$, so $x+y=2$ and $x^{4}+y^{4}=34$. We know $$(x+y)^{4}=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}=x^{4}+y^{4}+2 x y(2 x^{2}+2 y^{2}+3 x y) .$$ Moreover, $x^{2}+y^{2}=(x+y)^{2}-2 x y$, so the preceding equation becomes $2^{4}=34+2 x y(2. 2^{2}-x y)$, or $(x y)^{2}-8 x y-9=0$. Hence $x y=9$ or -1 . Solving $x y=9, x+y=2$ produces complex solutions, and solving $x y=-1, x+y=2$ produces $(x, y)=(1+\sqrt{2}, 1-\sqrt{2})$ or $(1-\sqrt{2}, 1+\sqrt{2})$. Thus, $x=1 \pm \sqrt{2}$.
|
1 \pm \sqrt{2}
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics",
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5
|
We wish to color the integers $1,2,3, \ldots, 10$ in red, green, and blue, so that no two numbers $a$ and $b$, with $a-b$ odd, have the same color. (We do not require that all three colors be used.) In how many ways can this be done?
|
The condition is equivalent to never having an odd number and an even number in the same color. We can choose one of the three colors for the odd numbers and distribute the other two colors freely among the 5 even numbers; this can be done in $3 \cdot 2^{5}=96$ ways. We can also choose one color for the even numbers and distribute the other two colors among the 5 odd numbers, again in 96 ways. This gives a total of 192 possibilities. However, we have double-counted the $3 \cdot 2=6$ cases where all odd numbers are the same color and all even numbers are the same color, so there are actually $192-6=186$ possible colorings.
|
186
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Graph Theory"
] | 5
|
Euler's Bridge: The following figure is the graph of the city of Konigsburg in 1736 - vertices represent sections of the cities, edges are bridges. An Eulerian path through the graph is a path which moves from vertex to vertex, crossing each edge exactly once. How many ways could World War II bombers have knocked out some of the bridges of Konigsburg such that the Allied victory parade could trace an Eulerian path through the graph? (The order in which the bridges are destroyed matters.)
|
The number of ways to destroy bridges to create an Eulerian path depends on ensuring that exactly 0 or 2 vertices have an odd degree. The specific graph of Konigsburg can be analyzed to find the number of such configurations, resulting in 13023 ways.
|
13023
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5
|
Compute: $$\left\lfloor\frac{2005^{3}}{2003 \cdot 2004}-\frac{2003^{3}}{2004 \cdot 2005}\right\rfloor$$
|
Let $x=2004$. Then the expression inside the floor brackets is $$\frac{(x+1)^{3}}{(x-1) x}-\frac{(x-1)^{3}}{x(x+1)}=\frac{(x+1)^{4}-(x-1)^{4}}{(x-1) x(x+1)}=\frac{8 x^{3}+8 x}{x^{3}-x}=8+\frac{16 x}{x^{3}-x}$$ Since $x$ is certainly large enough that $0<16 x /(x^{3}-x)<1$, the answer is 8.
|
8
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5
|
Find all ordered triples $(a, b, c)$ of positive reals that satisfy: $\lfloor a\rfloor b c=3, a\lfloor b\rfloor c=4$, and $a b\lfloor c\rfloor=5$, where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.
|
Write $p=a b c, q=\lfloor a\rfloor\lfloor b\rfloor\lfloor c\rfloor$. Note that $q$ is an integer. Multiplying the three equations gives: $$p=\sqrt{\frac{60}{q}}$$ Substitution into the first equation, $$p=3 \frac{a}{\lfloor a\rfloor}<3 \frac{\lfloor a\rfloor+1}{\lfloor a\rfloor} \leq 6$$ Looking at the last equation: $$p=5 \frac{c}{\lfloor c\rfloor} \geq 5 \frac{\lfloor c\rfloor}{\lfloor c\rfloor} \geq 5$$ Here we've used $\lfloor x\rfloor \leq x<\lfloor x\rfloor+1$, and also the apparent fact that $\lfloor a\rfloor \geq 1$. Now: $$\begin{gathered} 5 \leq \sqrt{\frac{60}{q}} \leq 6 \\ \frac{12}{5} \geq q \geq \frac{5}{3} \end{gathered}$$ Since $q$ is an integer, we must have $q=2$. Since $q$ is a product of 3 positive integers, we must have those be 1,1 , and 2 in some order, so there are three cases: Case 1: $\lfloor a\rfloor=2$. By the equations, we'd need $a=\frac{2}{3} \sqrt{30}=\sqrt{120 / 9}>3$, a contradiction, so there are no solutions in this case. Case 2: $\lfloor b\rfloor=2$. We have the solution $$\left(\frac{\sqrt{30}}{3}, \frac{\sqrt{30}}{2}, \frac{\sqrt{30}}{5}\right)$$ Case 3: $\lfloor c\rfloor=2$. We have the solution $$\left(\frac{\sqrt{30}}{3}, \frac{\sqrt{30}}{4}, \frac{2 \sqrt{30}}{5}\right)$$
|
\left(\frac{\sqrt{30}}{3}, \frac{\sqrt{30}}{4}, \frac{2 \sqrt{30}}{5}\right),\left(\frac{\sqrt{30}}{3}, \frac{\sqrt{30}}{2}, \frac{\sqrt{30}}{5}\right)
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Sequences and Series"
] | 5
|
An infinite sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the recurrence $$a_{n+3}=a_{n+2}-2 a_{n+1}+a_{n}$$ for every positive integer $n$. Given that $a_{1}=a_{3}=1$ and $a_{98}=a_{99}$, compute $a_{1}+a_{2}+\cdots+a_{100}$.
|
A quick telescope gives that $a_{1}+\cdots+a_{n}=2 a_{1}+a_{3}+a_{n-1}-a_{n-2}$ for all $n \geq 3$: $$\begin{aligned} \sum_{k=1}^{n} a_{k} & =a_{1}+a_{2}+a_{3}+\sum_{k=1}^{n-3}\left(a_{k}-2 a_{k+1}+2 a_{k+2}\right) \\ & =a_{1}+a_{2}+a_{3}+\sum_{k=1}^{n-3} a_{k}-2 \sum_{k=2}^{n-2} a_{k}+\sum_{k=3}^{n-1} a_{k} \\ & =2 a_{1}+a_{3}-a_{n-2}+a_{n-1} \end{aligned}$$ Putting $n=100$ gives the answer. One actual value of $a_{2}$ which yields the sequence is $a_{2}=\frac{742745601954}{597303450449}$.
|
3
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Precalculus -> Trigonometric Functions"
] | 5
|
How many real triples $(a, b, c)$ are there such that the polynomial $p(x)=x^{4}+a x^{3}+b x^{2}+a x+c$ has exactly three distinct roots, which are equal to $\tan y, \tan 2 y$, and $\tan 3 y$ for some real $y$ ?
|
Let $p$ have roots $r, r, s, t$. Using Vieta's on the coefficient of the cubic and linear terms, we see that $2 r+s+t=r^{2} s+r^{2} t+2 r s t$. Rearranging gives $2 r(1-s t)=\left(r^{2}-1\right)(s+t)$. If $r^{2}-1=0$, then since $r \neq 0$, we require that $1-s t=0$ for the equation to hold. Conversely, if $1-s t=0$, then since $s t=1, s+t=0$ cannot hold for real $s, t$, we require that $r^{2}-1=0$ for the equation to hold. So one valid case is where both these values are zero, so $r^{2}=s t=1$. If $r=\tan y$ (here we stipulate that $0 \leq y<\pi$ ), then either $y=\frac{\pi}{4}$ or $y=\frac{3 \pi}{4}$. In either case, the value of $\tan 2 y$ is undefined. If $r=\tan 2 y$, then we have the possible values $y=\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}$. In each of these cases, we must check if $\tan y \tan 3 y=1$. But this is true if $y+3 y=4 y$ is a odd integer multiple of $\frac{\pi}{2}$, which is the case for all such values. If $r=\tan 3 y$, then we must have $\tan y \tan 2 y=1$, so that $3 y$ is an odd integer multiple of $\frac{\pi}{2}$. But then $\tan 3 y$ would be undefined, so none of these values can work. Now, we may assume that $r^{2}-1$ and $1-s t$ are both nonzero. Dividing both sides by $\left(r^{2}-1\right)(1-s t)$ and rearranging yields $0=\frac{2 r}{1-r^{2}}+\frac{s+t}{1-s t}$, the tangent addition formula along with the tangent double angle formula. By setting $r$ to be one of $\tan y, \tan 2 y$, or $\tan 3 y$, we have one of the following: (a) $0=\tan 2 y+\tan 5 y$ (b) $0=\tan 4 y+\tan 4 y$ (c) $0=\tan 6 y+\tan 3 y$. We will find the number of solutions $y$ in the interval $[0, \pi)$. Case 1 yields six multiples of $\frac{\pi}{7}$. Case 2 yields $\tan 4 y=0$, which we can readily check has no solutions. Case 3 yields eight multiples of $\frac{\pi}{9}$. In total, we have $4+6+8=18$ possible values of $y$.
|
18
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Probability -> Other"
] | 5
|
Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a $60 \%$ chance of winning each point, what is the probability that he will win the game?
|
Consider the situation after two points. Daniel has a $9 / 25$ chance of winning, Scott, $4 / 25$, and there is a $12 / 25$ chance that the players will be tied. In the latter case, we revert to the original situation. In particular, after every two points, either the game returns to the original situation, or one player wins. If it is given that the game lasts $2 k$ rounds, then the players must be at par after $2(k-1)$ rounds, and then Daniel wins with probability $(9 / 25) /(9 / 25+4 / 25)=9 / 13$. Since this holds for any $k$, we conclude that Daniel wins the game with probability $9 / 13$.
|
9 / 13
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5
|
Find the unique pair of positive integers $(a, b)$ with $a<b$ for which $$\frac{2020-a}{a} \cdot \frac{2020-b}{b}=2$$
|
If either $a$ or $b$ is larger than 2020, then both must be for the product to be positive. However, the resulting product would be less than 1, so this case is impossible. Now, we see that $\left(\frac{2020-a}{a}, \frac{2020-b}{b}\right)$ must be in the form $\left(\frac{x}{y}, \frac{2 y}{x}\right)$, in some order, for relatively prime positive integers $x$ and $y$. Then $\frac{2020}{a}=\frac{x+y}{y}$ and $\frac{2020}{b}=\frac{x+2 y}{x}$, so $x+y$ and $x+2 y$ are relatively prime factors of 2020. Since $x+y<x+2 y<2(x+y)$, the only possibility is $x+y=4, x+2 y=5$. Thus, $(x, y)=(3,1)$, and $\left(\frac{2020-a}{a}, \frac{2020-b}{b}\right)=\left(3, \frac{2}{3}\right)$ because $a<b$. Solving gives $(a, b)=(505,1212)$.
|
(505,1212)
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions",
"Mathematics -> Precalculus -> Trigonometric Functions"
] | 5
|
Compute $\frac{\tan ^{2}\left(20^{\circ}\right)-\sin ^{2}\left(20^{\circ}\right)}{\tan ^{2}\left(20^{\circ}\right) \sin ^{2}\left(20^{\circ}\right)}$.
|
If we multiply top and bottom by $\cos ^{2}\left(20^{\circ}\right)$, the numerator becomes $\sin ^{2}\left(20^{\circ}\right) \cdot(1-\cos ^{2} 20^{\circ})=\sin ^{4}\left(20^{\circ}\right)$, while the denominator becomes $\sin ^{4}\left(20^{\circ}\right)$ also. So they are equal, and the ratio is 1.
|
1
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Number Theory -> Factorization"
] | 5
|
Let $S$ be a set of positive integers satisfying the following two conditions: - For each positive integer $n$, at least one of $n, 2 n, \ldots, 100 n$ is in $S$. - If $a_{1}, a_{2}, b_{1}, b_{2}$ are positive integers such that $\operatorname{gcd}\left(a_{1} a_{2}, b_{1} b_{2}\right)=1$ and $a_{1} b_{1}, a_{2} b_{2} \in S$, then $a_{2} b_{1}, a_{1} b_{2} \in S$ Suppose that $S$ has natural density $r$. Compute the minimum possible value of $\left\lfloor 10^{5} r\right\rfloor$. Note: $S$ has natural density $r$ if $\frac{1}{n}|S \cap\{1, \ldots, n\}|$ approaches $r$ as $n$ approaches $\infty$.
|
The optimal value of $r$ is $\frac{1}{252}$. This is attained by letting $S$ be the set of integers $n$ for which $\nu_{2}(n) \equiv 4 \bmod 5$ and $\nu_{3}(n) \equiv 1 \bmod 2$. Let $S$ be a set of positive integers satisfying the two conditions. For each prime $p$, let $A_{p}=\left\{\nu_{p}(n)\right.$ : $n \in S\}$. We claim that in fact $S$ is precisely the set of positive integers $n$ for which $\nu_{p}(n) \in A_{p}$ for each prime $p$. Let $p$ be prime and suppose that $a_{1} p^{e_{1}}, a_{2} p^{e_{2}} \in S$, with $p \nmid a_{1}, a_{2}$. Then, setting $b_{1}=p^{e_{1}}$ and $b_{2}=p^{e_{2}}$ in the second condition gives that $a_{1} p^{e_{2}} \in S$ as well. So, if we have an integer $n$ for which $\nu_{p}(n) \in A_{p}$ for each prime $p$, we can start with any element $n^{\prime}$ of $S$ and apply this step for each prime divisor of $n$ and $n^{\prime}$ to obtain $n \in S$. Now we deal with the first condition. Let $n$ be any positive integer. We will compute the least positive integer $m$ such that $m n \in S$. By the above result, we can work with each prime separately. For a given prime $p$, let $e_{p}$ be the least element of $A_{p}$ with $e_{p} \geq \nu_{p}(n)$. Then we must have $\nu_{p}(m) \geq e_{p}-\nu_{p}(n)$, and equality for all primes $p$ is sufficient. So, if the elements of $A_{p}$ are $c_{p, 1}<c_{p, 2}<c_{p, 3}<c_{p, 4}<\ldots$, then $$c_{p}=\max \left(c_{p, 1}, c_{p, 2}-c_{p, 1}-1, c_{p, 3}-c_{p, 2}-1, c_{p, 4}-c_{p, 3}-1, \ldots\right)$$ is the worst case value for $\nu_{p}(m)$. We conclude two things from this. First, we must have $\prod_{p} p^{c_{p}} \leq 100$ by condition 1, and in fact this is sufficient. Second, since we only care about $c_{p}$ and would like to minimize $r$, the optimal choice for $A_{p}$ is an arithmetic progression with first term $c_{p}$ and common difference $c_{p}+1$. So we assume that each $A_{p}$ is of this form. Let $t=\prod_{p} p^{c_{p}}$. We now compute $r$. Note that $S$ is the set of integers $n$ such that for each prime $p$, $$n \equiv a p^{k\left(c_{p}+1\right)-1} \bmod p^{k\left(c_{p}+1\right)}$$ for some positive integers $a, k$ with $a<p$. This means that each prime $p$ contributes a factor of $$\frac{p-1}{p^{c_{p}+1}}+\frac{p-1}{p^{2 c_{p}+2}}+\frac{p-1}{p^{3 c_{p}+3}}+\cdots=\frac{p-1}{p^{c_{p}+1}-1}=\frac{1}{1+p+\cdots+p^{c_{p}}}$$ to the density of $S$. Multiplying over all primes $p$ gives $r=\frac{1}{\sigma(t)}$, where $\sigma(t)$ is the sum of divisors of $t$. So, it suffices to maximize $\sigma(t)$ for $t \leq 100$. By inspection, $t=96$ is optimal, giving $r=\frac{1}{252}$.
|
396
|
HMMT_2
|
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5
|
In triangle $A B C, \angle A B C$ is obtuse. Point $D$ lies on side $A C$ such that \angle A B D$ is right, and point $E$ lies on side $A C$ between $A$ and $D$ such that $B D$ bisects \angle E B C$. Find $C E$, given that $A C=35, B C=7$, and $B E=5$.
|
Reflect $A$ and $E$ over $B D$ to $A^{\prime}$ and $E^{\prime}$ respectively. Note that the angle conditions show that $A^{\prime}$ and $E^{\prime}$ lie on $A B$ and $B C$ respectively. $B$ is the midpoint of segment $A A^{\prime}$ and $C E^{\prime}=$ $B C-B E^{\prime}=2$. Menelaus' theorem now gives $$\frac{C D}{D A} \cdot \frac{A A^{\prime}}{A^{\prime} B} \cdot \frac{B E^{\prime}}{E^{\prime} C}=1$$ from which $D A=5 C D$ or $C D=A C / 6$. By the angle bisector theorem, $D E=5 C D / 7$, so that $C E=12 C D / 7=10$.
|
10
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Factorization",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5
|
For any positive integer $n$, let $\tau(n)$ denote the number of positive divisors of $n$. If $n$ is a positive integer such that $\frac{\tau\left(n^{2}\right)}{\tau(n)}=3$, compute $\frac{\tau\left(n^{7}\right)}{\tau(n)}$.
|
Let the prime factorization of $n$ be $n=p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{k}^{e_{k}}$. Then, the problem condition is equivalent to $$\prod_{i=1}^{k} \frac{2 e_{i}+1}{e_{i}+1}=3$$ Note that since $\frac{2 x+1}{x+1} \geq 1.5$ for $x \geq 1$, and $1.5^{3}>3$, we have $k \leq 2$. Also, $k=1$ implies $2 e_{1}+1=3\left(e_{1}+1\right)$, which implies $e_{1}$ is negative. Thus, we must have $k=2$. Then, our equation becomes $$\left(2 e_{1}+1\right)\left(2 e_{2}+1\right)=3\left(e_{1}+1\right)\left(e_{2}+1\right)$$ which simplifies to $\left(e_{1}-1\right)\left(e_{2}-1\right)=3$. This gives us $e_{1}=2$ and $e_{2}=4$. Thus, we have $n=p^{2} q^{4}$ for primes $p$ and $q$, so $\frac{\tau\left(n^{7}\right)}{\tau(n)}=\frac{\tau\left(p^{14} q^{28}\right)}{\tau\left(p^{2} q^{4}\right)}=\frac{15 \cdot 29}{3 \cdot 5}=29$.
|
29
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Bob is writing a sequence of letters of the alphabet, each of which can be either uppercase or lowercase, according to the following two rules: If he had just written an uppercase letter, he can either write the same letter in lowercase after it, or the next letter of the alphabet in uppercase. If he had just written a lowercase letter, he can either write the same letter in uppercase after it, or the preceding letter of the alphabet in lowercase. For instance, one such sequence is $a A a A B C D d c b B C$. How many sequences of 32 letters can he write that start at (lowercase) $a$ and end at (lowercase) $z$?
|
The smallest possible sequence from $a$ to $z$ is $a A B C D \ldots Z z$, which has 28 letters. To insert 4 more letters, we can either switch two (not necessarily distinct) letters to lowercase and back again (as in $a A B C c C D E F f F G H \ldots Z z$ ), or we can insert a lowercase letter after its corresponding uppercase letter, insert the previous letter of the alphabet, switch back to uppercase, and continue the sequence (as in $a A B C c b B C D E \ldots Z z$ ). There are $\binom{27}{2}=13 \cdot 27$ sequences of the former type and 25 of the latter, for a total of 376 such sequences.
|
376
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Other"
] | 5
|
During the regular season, Washington Redskins achieve a record of 10 wins and 6 losses. Compute the probability that their wins came in three streaks of consecutive wins, assuming that all possible arrangements of wins and losses are equally likely. (For example, the record LLWWWWWLWWLWWWLL contains three winning streaks, while WWWWWWWLLLLLLWWW has just two.)
|
Suppse the winning streaks consist of $w_{1}, w_{2}$, and $w_{3}$ wins, in chronological order, where the first winning streak is preceded by $l_{0}$ consecutive losses and the $i$ winning streak is immediately succeeded by $l_{i}$ losses. Then $w_{1}, w_{2}, w_{3}, l_{1}, l_{2}>0$ are positive and $l_{0}, l_{3} \geq 0$ are nonnegative. The equations $$w_{1}+w_{2}+w_{3}=10 \quad \text { and } \quad\left(l_{0}+1\right)+l_{1}+l_{2}+\left(l_{3}+1\right)=8$$ are independent, and have $\binom{9}{2}$ and $\binom{7}{3}$ solutions, respectively. It follows that the answer is $$\frac{\binom{9}{2}\binom{7}{3}}{\binom{16}{6}}=\frac{315}{2002}$$
|
\frac{315}{2002}
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
In how many distinct ways can you color each of the vertices of a tetrahedron either red, blue, or green such that no face has all three vertices the same color? (Two colorings are considered the same if one coloring can be rotated in three dimensions to obtain the other.)
|
If only two colors are used, there is only one possible arrangement up to rotation, so this gives 3 possibilities. If all three colors are used, then one is used twice. There are 3 ways to choose the color that is used twice. Say this color is red. Then the red vertices are on a common edge, and the green and blue vertices are on another edge. We see that either choice of arrangement of the green and blue vertices is the same up to rotation. Thus there are 6 possibilities total.
|
6
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Prime Numbers"
] | 5
|
The average of a set of distinct primes is 27. What is the largest prime that can be in this set?
|
Denote the set of these primes by $A$ and the number of elements in $A$ by n. There are 9 primes smaller than 27, namely $2,3,5,7,11,13,17,19$ and 23. Since 27 is odd and all primes except 2 are odd, $2 \notin A$. Thus the largest prime $p$ is at most $27 \cdot 9-3-5-7-11-13-17-19-23=145$, so $p \leq 141$. When the primes are $3,5,7,11,13,17,19,23,29,31$ and 139, their average is 27. Therefore $p=139$.
|
139
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5
|
Sam spends his days walking around the following $2 \times 2$ grid of squares. Say that two squares are adjacent if they share a side. He starts at the square labeled 1 and every second walks to an adjacent square. How many paths can Sam take so that the sum of the numbers on every square he visits in his path is equal to 20 (not counting the square he started on)?
|
Note that on the first step, Sam can either step on 2 or 4. On the second step, Sam can either step on 1 or 3, regardless of whether he is on 2 or 4. Now, for example, say that Sam takes 8 steps. His total sum will be $2+1+2+1+2+1+2+1+2 a$, where $a$ is the number of times that he decides to step on the larger number of his two choices. Solving gives $a=4$. As he took 8 steps, this gives him $\binom{8}{4}=70$ ways in this case. We can follow a similar approach by doing casework on the number of steps he takes. I will simply list them out here for brevity. For 8 steps, we get $\binom{8}{4}=70$. For 9 steps, we get $\binom{9}{3}=84$. For 12 steps, we get a contribution on $\binom{12}{1}=12$. For 13 steps, we get a contribution of $\binom{13}{0}=1$. Therefore, the final answer is $70+84+12+1=167$.
|
167
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions"
] | 5
|
Find all pairs of integer solutions $(n, m)$ to $2^{3^{n}}=3^{2^{m}}-1$.
|
We find all solutions of $2^{x}=3^{y}-1$ for positive integers $x$ and $y$. If $x=1$, we obtain the solution $x=1, y=1$, which corresponds to $(n, m)=(0,0)$ in the original problem. If $x>1$, consider the equation modulo 4. The left hand side is 0, and the right hand side is $(-1)^{y}-1$, so $y$ is even. Thus we can write $y=2 z$ for some positive integer $z$, and so $2^{x}=(3^{z}-1)(3^{z}+1)$. Thus each of $3^{z}-1$ and $3^{z}+1$ is a power of 2, but they differ by 2, so they must equal 2 and 4 respectively. Therefore, the only other solution is $x=3$ and $y=2$, which corresponds to $(n, m)=(1,1)$ in the original problem.
|
(0,0) \text{ and } (1,1)
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
The number $$316990099009901=\frac{32016000000000001}{101}$$ is the product of two distinct prime numbers. Compute the smaller of these two primes.
|
Let $x=2000$, so the numerator is $$x^{5}+x^{4}+1=\left(x^{2}+x+1\right)\left(x^{3}-x+1\right)$$ (This latter factorization can be noted by the fact that plugging in $\omega$ or $\omega^{2}$ into $x^{5}+x^{4}+1$ gives 0 .) Then $x^{2}+x+1=4002001$ divides the numerator. However, it can easily by checked that 101 doesn't divide 4002001 (since, for example, $101 \nmid 1-20+0-4$ ), so 4002001 is one of the primes. Then the other one is $$\frac{2000^{3}-2000+1}{101} \approx \frac{2000^{3}}{101}>2000^{2} \approx 4002001$$ so 4002001 is the smaller of the primes.
|
4002001
|
HMMT_2
|
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5
|
Let $A B C D$ be a convex quadrilateral such that $\angle A B D=\angle B C D=90^{\circ}$, and let $M$ be the midpoint of segment $B D$. Suppose that $C M=2$ and $A M=3$. Compute $A D$.
|
Since triangle $B C D$ is a right triangle, we have $C M=B M=D M=2$. With $A M=3$ and $\angle A B M=90^{\circ}$, we get $A B=\sqrt{5}$. Now $$A D^{2}=A B^{2}+B D^{2}=5+16=21$$ so $A D=\sqrt{21}$.
|
\sqrt{21}
|
HMMT_2
|
[
"Mathematics -> Algebra -> Abstract Algebra -> Other",
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 5
|
The vertices of a regular hexagon are labeled $\cos (\theta), \cos (2 \theta), \ldots, \cos (6 \theta)$. For every pair of vertices, Bob draws a blue line through the vertices if one of these functions can be expressed as a polynomial function of the other (that holds for all real $\theta$ ), and otherwise Roberta draws a red line through the vertices. In the resulting graph, how many triangles whose vertices lie on the hexagon have at least one red and at least one blue edge?
|
The existence of the Chebyshev polynomials, which express $\cos (n \theta)$ as a polynomial in $\cos (\theta)$, imply that Bob draws a blue line between $\cos (\theta)$ and each other vertex, and also between $\cos (2 \theta)$ and $\cos (4 \theta)$, between $\cos (2 \theta)$ and $\cos (6 \theta)$, and between $\cos (3 \theta)$ and $\cos (6 \theta)$ (by substituting $\theta^{\prime}=2 \theta$ or $3 \theta$ as necessary). We now show that Roberta draws a red line through each other pair of vertices. Let $m$ and $n$ be positive integers. Notice that $\cos (n \theta)$ is a periodic function with period $\frac{2 \pi}{n}$, and $\cos (m \theta)$ is periodic with period $\frac{2 \pi}{m}$. Thus, any polynomial in $\cos (m \theta)$ is also periodic of period $\frac{2 \pi}{m}$. This may not be the minimum period of the polynomial, however, so the minimum period is $\frac{2 \pi}{m k}$ for some $k$. Therefore, if $\cos (n \theta)$ can be expressed as a polynomial in $\cos (m \theta)$ then $\frac{2 \pi}{n}=\frac{2 \pi}{m k}$ for some $k$, so $m \mid n$. This shows that there is a blue line between two vertices $\cos (a \theta)$ and $\cos (b \theta)$ if and only if one of $a$ or $b$ divides the other. Drawing the graph, one can easily count that there are 3 triangles with all blue edges, 3 triangles with all red edges, and $\binom{6}{3}=20$ triangles total. Thus there are $20-3-3=14$ triangles having at least one red and at least one blue edge.
|
14
|
HMMT_2
|
[
"Mathematics -> Algebra -> Intermediate Algebra -> Other"
] | 5
|
Let $a_{1}=1$, and let $a_{n}=\left\lfloor n^{3} / a_{n-1}\right\rfloor$ for $n>1$. Determine the value of $a_{999}$.
|
We claim that for any odd $n, a_{n}=n$. The proof is by induction. To get the base cases $n=1$, 3, we compute $a_{1}=1, a_{2}=\left\lfloor 2^{3} / 1\right\rfloor=8, a_{3}=\left\lfloor 3^{3} / 8\right\rfloor=3$. And if the claim holds for odd $n \geq 3$, then $a_{n+1}=\left\lfloor(n+1)^{3} / n\right\rfloor=n^{2}+3 n+3$, so $a_{n+2}=\left\lfloor(n+2)^{3} /\left(n^{2}+3 n+3\right)\right\rfloor=\left\lfloor\left(n^{3}+6 n^{2}+12 n+8\right) /\left(n^{2}+3 n+2\right)\right\rfloor=\left\lfloor n+2+\frac{n^{2}+3 n+2}{n^{2}+3 n+3}\right\rfloor=n+2$. So the claim holds, and in particular, $a_{999}=999$.
|
999
|
HMMT_2
|
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations",
"Mathematics -> Geometry -> Plane Geometry -> Circles"
] | 5
|
$A B C D$ is a cyclic quadrilateral in which $A B=3, B C=5, C D=6$, and $A D=10 . M, I$, and $T$ are the feet of the perpendiculars from $D$ to lines $A B, A C$, and $B C$ respectively. Determine the value of $M I / I T$.
|
Quadrilaterals $A M I D$ and $D I C T$ are cyclic, having right angles $\angle A M D, \angle A I D$, and $\angle C I D, \angle C T D$ respectively. We see that $M, I$, and $T$ are collinear. For, $m \angle M I D=\pi-m \angle D A M=$ $\pi-m \angle D A B=m \angle B C D=\pi-m \angle D C T=\pi-m \angle D I T$. Therefore, Menelaus' theorem applied to triangle MTB and line $I C A$ gives $$\frac{M I}{I T} \cdot \frac{T C}{C B} \cdot \frac{B A}{A M}=1$$ On the other hand, triangle $A D M$ is similar to triangle $C D T$ since $\angle A M D \cong \angle C T D$ and $\angle D A M \cong$ $\angle D C T$ and thus $A M / C T=A D / C D$. It follows that $$\frac{M I}{I T}=\frac{B C \cdot A M}{A B \cdot C T}=\frac{B C \cdot A D}{A B \cdot C D}=\frac{5 \cdot 10}{3 \cdot 6}=\frac{25}{9}$$ Remarks. The line $M I T$, constructed in this problem by taking perpendiculars from a point on the circumcircle of $A B C$, is known as the Simson line. It is often helpful for us to use directed angles while angle chasing to avoid supplementary configuration issues, such as those arising while establishing the collinearity of $M, I$, and $T$.
|
\frac{25}{9}
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Sequences and Series"
] | 5
|
Let $a_{1}, a_{2}, \ldots$ be a sequence defined by $a_{1}=a_{2}=1$ and $a_{n+2}=a_{n+1}+a_{n}$ for $n \geq 1$. Find $$\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}$$
|
Let $X$ denote the desired sum. Note that $$\begin{array}{rlr} X & = & \frac{1}{4^{2}}+\frac{1}{4^{3}}+\frac{2}{4^{4}}+\frac{3}{4^{5}}+\frac{5}{4^{6}}+\ldots \\ 4 X & =\quad \frac{1}{4^{1}}+\frac{1}{4^{2}}+\frac{2}{4^{3}}+\frac{3}{4^{4}}+\frac{5}{4^{5}}+\frac{8}{4^{6}}+\ldots \\ 16 X & =\frac{1}{4^{0}}+\frac{1}{4^{1}}+\frac{2}{4^{2}}+\frac{3}{4^{3}}+\frac{5}{4^{4}}+\frac{8}{4^{5}}+\frac{13}{4^{6}}+\ldots \end{array}$$ so that $X+4 X=16 X-1$, and $X=1 / 11$.
|
\frac{1}{11}
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Other"
] | 5
|
Bob the bomb-defuser has stumbled upon an active bomb. He opens it up, and finds the red and green wires conveniently located for him to cut. Being a seasoned member of the bomb-squad, Bob quickly determines that it is the green wire that he should cut, and puts his wirecutters on the green wire. But just before he starts to cut, the bomb starts to count down, ticking every second. Each time the bomb ticks, starting at time $t=15$ seconds, Bob panics and has a certain chance to move his wirecutters to the other wire. However, he is a rational man even when panicking, and has a $\frac{1}{2 t^{2}}$ chance of switching wires at time $t$, regardless of which wire he is about to cut. When the bomb ticks at $t=1$, Bob cuts whatever wire his wirecutters are on, without switching wires. What is the probability that Bob cuts the green wire?
|
Suppose Bob makes $n$ independent decisions, with probabilities of switching $p_{1}, p_{2}, \ldots, p_{n}$. Then in the expansion of the product $$P(x)=\left(p_{1}+\left(1-p_{1}\right) x\right)\left(p_{2}+\left(1-p_{2}\right) x\right) \cdots\left(p_{n}+\left(1-p_{n}\right) x\right)$$ the sum of the coefficients of even powers of $x$ gives the probability that Bob makes his original decision. This is just $(P(1)+P(-1)) / 2$, so the probability is just $$\frac{1+\left(1-\frac{1}{1515}\right)\left(1-\frac{1}{1414}\right) \cdots\left(1-\frac{1}{22}\right)}{2}=\frac{1+\frac{1416}{15151315} \cdots \frac{13}{22}}{2}=\frac{1+\frac{8}{15}}{2}=\frac{23}{30}$$
|
\frac{23}{30}
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Other"
] | 5
|
Given a positive integer $k$, let \|k\| denote the absolute difference between $k$ and the nearest perfect square. For example, \|13\|=3 since the nearest perfect square to 13 is 16. Compute the smallest positive integer $n$ such that $\frac{\|1\|+\|2\|+\cdots+\|n\|}{n}=100$.
|
Note that from $n=m^{2}$ to $n=(m+1)^{2},\|n\|$ increases from 0 to a peak of $m$ (which is repeated twice), and then goes back down to 0. Therefore $\sum_{n=1}^{m^{2}}\|n\|=\sum_{k=1}^{m-1} 2(1+2+\cdots+k)=\sum_{k=1}^{m-1} 2\binom{k+1}{2}=2\binom{m+1}{3}=\frac{m}{3}\left(m^{2}-1\right)$. In particular, if $n=m^{2}-1$, $\frac{\|1\|+\|2\|+\cdots+\|n\|}{n}=\frac{m}{3}$ so $n=300^{2}-1$ satisfies the condition. However, this does not prove that there are not smaller solutions for $n$. Let $N=300^{2}-1$ and suppose that $N-k$ satisfies the condition. Then, we know that $\frac{\|N\|+\|N-1\|+\cdots\|N-(k-1)\|}{k}=100$. Since \|N-k\|=k+1 for $k \leq 298$, one can show that $k=199$ works. By looking at further terms, one can convince oneself that no larger value of $k$ works. Thus, the answer is $300^{2}-1-199=90000-200=$ 89800.
|
89800
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5
|
Find the number of 7 -tuples $\left(n_{1}, \ldots, n_{7}\right)$ of integers such that $$\sum_{i=1}^{7} n_{i}^{6}=96957$$
|
Consider the equation in modulo 9. All perfect 6 th powers are either 0 or 1. Since 9 divides 96957, it must be that each $n_{i}$ is a multiple of 3. Writing $3 a_{i}=n_{i}$ and dividing both sides by $3^{6}$, we have $a_{1}^{6}+\cdots+a_{7}^{6}=133$. Since sixth powers are nonnegative, $\left|a_{i}\right| \leq 2$. Again considering modulo 9, we see that $a_{i} \neq 0$. Thus, $a_{i}^{6} \in\{1,64\}$. The only possibility is $133=64+64+1+1+1+1+1$, so $\left|a_{1}\right|, \ldots,\left|a_{7}\right|$ consists of 22 's and 51 's. It follows that the answer is $\binom{7}{2} \cdot 2^{7}=2688$.
|
2688
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Compute, in terms of $n$, $\sum_{k=0}^{n}\binom{n-k}{k} 2^{k}$.
|
Let $T_{n}=\sum_{k=0}^{n}\binom{n-k}{k} 2^{k}$. From Pascal's recursion for binomial coefficients, we can find $T_{n}=2 T_{n-2}+T_{n-1}$, with $T_{0}=1$ and $T_{1}=1$. The characteristic polynomial of this recursion is $x^{2}-x-2=0$, which has roots 2 and -1. Thus $T_{n}=a \cdot 2^{n}+b \cdot(-1)^{n}$ for some $a$ and $b$. From the initial conditions we have $a+b=1$ and $2 a-b=1$. It follows that $a=2 / 3$ and $b=1 / 3$, from which the conclusion follows.
|
\frac{2 \cdot 2^{n}+(-1)^{n}}{3}
|
HMMT_2
|
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Plane Geometry -> Angles"
] | 5
|
$A B C D$ is a cyclic quadrilateral in which $A B=4, B C=3, C D=2$, and $A D=5$. Diagonals $A C$ and $B D$ intersect at $X$. A circle $\omega$ passes through $A$ and is tangent to $B D$ at $X . \omega$ intersects $A B$ and $A D$ at $Y$ and $Z$ respectively. Compute $Y Z / B D$.
|
Denote the lengths $A B, B C, C D$, and $D A$ by $a, b, c$, and $d$ respectively. Because $A B C D$ is cyclic, $\triangle A B X \sim \triangle D C X$ and $\triangle A D X \sim \triangle B C X$. It follows that $\frac{A X}{D X}=\frac{B X}{C X}=\frac{a}{c}$ and $\frac{A X}{B X}=\frac{D X}{C X}=\frac{d}{b}$. Therefore we may write $A X=a d k, B X=a b k, C X=b c k$, and $D X=c d k$ for some $k$. Now, $\angle X D C=\angle B A X=\angle Y X B$ and $\angle D C X=\angle X B Y$, so $\triangle B X Y \sim \triangle C D X$. Thus, $X Y=$ $D X \cdot \frac{B X}{C D}=a b d k^{2}$. Analogously, $X Y=a c d k^{2}$. Note that $X Y / X Z=C B / C D$. Since $\angle Y X Z=\pi-\angle Z A \stackrel{c}{Y}=\angle B C D$, we have that $\triangle X Y Z \sim \triangle C B D$. Thus, $Y Z / B D=X Y / C B=a d k^{2}$. Finally, Ptolemy's theorem applied to $A B C D$ gives $$(a d+b c) k \cdot(a b+c d) k=a c+b d$$ It follows that the answer is $$\frac{a d(a c+b d)}{(a b+c d)(a d+b c)}=\frac{20 \cdot 23}{22 \cdot 26}=\frac{115}{143}$$
|
\frac{115}{143}
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Other"
] | 5
|
Descartes's Blackjack: How many integer lattice points (points of the form $(m, n)$ for integers $m$ and $n$) lie inside or on the boundary of the disk of radius 2009 centered at the origin?
|
The number of lattice points inside or on the boundary of a circle with radius $r$ centered at the origin can be approximated using the formula $\pi r^2 + \text{error term}$. For a circle with radius 2009, this results in approximately 12679605 lattice points.
|
12679605
|
HMMT_2
|
[
"Mathematics -> Calculus -> Integral Calculus -> Techniques of Integration -> Single-variable",
"Mathematics -> Algebra -> Algebraic Expressions -> Other"
] | 5
|
Compute $$\sum_{n=0}^{\infty} \frac{n}{n^{4}+n^{2}+1}$$
|
Note that $$n^{4}+n^{2}+1=\left(n^{4}+2 n^{2}+1\right)-n^{2}=\left(n^{2}+1\right)^{2}-n^{2}=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)$$ Decomposing into partial fractions, we find that $$\frac{n}{n^{4}+n^{2}+1}=\frac{1}{2}\left(\frac{1}{n^{2}-n+1}-\frac{1}{n^{2}+n+1}\right)$$ Now, note that if $f(n)=\frac{1}{n^{2}-n+1}$, then $f(n+1)=\frac{1}{(n+1)^{2}-(n+1)+1}=\frac{1}{n^{2}+n+1}$. It follows that $$\sum_{n=0}^{\infty} \frac{n}{n^{4}+n^{2}+1}=\frac{1}{2}((f(0)-f(1))+(f(1)-f(2))+(f(2)-f(3))+\cdots)$$ Since $f(n)$ tends towards 0 as $n$ gets large, this sum telescopes to $f(0) / 2=1 / 2$.
|
1/2
|
HMMT_2
|
[
"Mathematics -> Precalculus -> Functions"
] | 5
|
Given that \(x\) is a positive real, find the maximum possible value of \(\sin \left(\tan ^{-1}\left(\frac{x}{9}\right)-\tan ^{-1}\left(\frac{x}{16}\right)\right)\).
|
Consider a right triangle \(A O C\) with right angle at \(O, A O=16\) and \(C O=x\). Moreover, let \(B\) be on \(A O\) such that \(B O=9\). Then \(\tan ^{-1} \frac{x}{9}=\angle C B O\) and \(\tan ^{-1} \frac{x}{16}=\angle C A O\), so their difference is equal to \(\angle A C B\). Note that the locus of all possible points \(C\) given the value of \(\angle A C B\) is part of a circle that passes through \(A\) and \(B\), and if we want to maximize this angle then we need to make this circle as small as possible. This happens when \(O C\) is tangent to the circumcircle of \(A B C\), so \(O C^{2}=O A \cdot O B=144=12^{2}\), thus \(x=12\), and it suffices to compute \(\sin (\alpha-\beta)\) where \(\sin \alpha=\cos \beta=\frac{4}{5}\) and \(\cos \alpha=\sin \beta=\frac{3}{5}\). By angle subtraction formula we get \(\sin (\alpha-\beta)=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}=\frac{7}{25}\).
|
\frac{7}{25}
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5
|
Two real numbers $x$ and $y$ are such that $8 y^{4}+4 x^{2} y^{2}+4 x y^{2}+2 x^{3}+2 y^{2}+2 x=x^{2}+1$. Find all possible values of $x+2 y^{2}$.
|
Writing $a=x+2 y^{2}$, the given quickly becomes $4 y^{2} a+2 x^{2} a+a+x=x^{2}+1$. We can rewrite $4 y^{2} a$ for further reduction to $a(2 a-2 x)+2 x^{2} a+a+x=x^{2}+1$, or $$\begin{equation*} 2 a^{2}+\left(2 x^{2}-2 x+1\right) a+\left(-x^{2}+x-1\right)=0 \tag{*} \end{equation*}$$ The quadratic formula produces the discriminant $$\left(2 x^{2}-2 x+1\right)^{2}+8\left(x^{2}-x+1\right)=\left(2 x^{2}-2 x+3\right)^{2}$$ an identity that can be treated with the difference of squares, so that $a=\frac{-2 x^{2}+2 x-1 \pm\left(2 x^{2}-2 x+3\right)}{4}=$ $\frac{1}{2},-x^{2}+x-1$. Now $a$ was constructed from $x$ and $y$, so is not free. Indeed, the second expression flies in the face of the trivial inequality: $a=-x^{2}+x-1<-x^{2}+x \leq x+2 y^{2}=a$. On the other hand, $a=1 / 2$ is a bona fide solution to $\left(^{*}\right)$, which is identical to the original equation.
|
\frac{1}{2}
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
The number 770 is written on a blackboard. Melody repeatedly performs moves, where a move consists of subtracting either 40 or 41 from the number on the board. She performs moves until the number is not positive, and then she stops. Let $N$ be the number of sequences of moves that Melody could perform. Suppose $N=a \cdot 2^{b}$ where $a$ is an odd positive integer and $b$ is a nonnegative integer. Compute $100 a+b$.
|
Notice that if we use the 41 move nine times or less, we will have to make a total of $\left\lceil\frac{770}{40}\right\rceil=20$ moves, and if we use it ten times or more, we will have to make a total of $\left\lfloor\frac{770}{40}\right\rfloor=19$ moves. So, doing casework on the number of 40 s we use gives $$\underbrace{\binom{19}{0}+\binom{19}{1}+\binom{19}{2}+\cdots+\binom{19}{9}}_{19 \text { moves }}+\underbrace{\frac{\binom{20}{10}}{2}+\binom{20}{11}+\binom{20}{11}+\cdots+\binom{20}{20}}_{20 \text { moves }}$$ Using the row sums of Pascal's triangle we have this sum equal to $\frac{2^{19}}{2}+\frac{2^{20}}{2}=3 \cdot 2^{18}$. The answer is 318.
|
318
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Factorization"
] | 5
|
How many perfect squares divide $2^{3} \cdot 3^{5} \cdot 5^{7} \cdot 7^{9}$?
|
The number of such perfect squares is $2 \cdot 3 \cdot 4 \cdot 5$, since the exponent of each prime can be any nonnegative even number less than the given exponent.
|
120
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions",
"Mathematics -> Algebra -> Linear Algebra -> Vectors"
] | 5
|
Kevin starts with the vectors \((1,0)\) and \((0,1)\) and at each time step, he replaces one of the vectors with their sum. Find the cotangent of the minimum possible angle between the vectors after 8 time steps.
|
Say that the vectors Kevin has at some step are \((a, b)\) and \((c, d)\). Notice that regardless of which vector he replaces with \((a+c, b+d)\), the area of the triangle with vertices \((0,0),(a, b)\), and \((c, d)\) is preserved with the new coordinates. We can see this geometrically: the parallelogram with vertices \((0,0),(a, b)\), \((c, d)\), and \((a+c, b+d)\) can be cut in half by looking at the triangle formed by any 3 of the vertices, which include the original triangle, and both possible triangles that might arise in the next step. Because the area is preserved, the minimum possible angle then arises when the two vectors, our sides, are as long as possible. This occurs when we alternate which vector is getting replaced for the sum. Given two vectors \((a, b)\) and \((c, d)\), with \(\sqrt{a^{2}+b^{2}}>\sqrt{c^{2}+d^{2}}\), we would rather replace \((c, d)\) than \((a, b)\), and \((a+c, b+d)\) has a larger norm than \((a, b)\). Then at the \(n\)th step, Kevin has the vectors \(\left(F_{n}, F_{n-1}\right)\) and \(\left(F_{n+1}, F_{n}\right)\), where \(F_{0}=0\) and \(F_{1}=1\). The tangent of the angle between them is the tangent of the difference of the angles they make with the x-axis, which is just their slope. We can then compute the cotangent as \(\left|\frac{1+\frac{F_{n-1}}{F_{n}} \cdot \frac{F_{n}}{F_{n+1}}}{\frac{F_{n}}{F_{n+1}}-\frac{F_{n-1}}{F_{n}}}\right|=\left|\frac{F_{n}\left(F_{n+1}+F_{n-1}\right)}{F_{n}^{2}-F_{n-1} F_{n+1}}\right|\). We can show (by induction) that \(F_{n}\left(F_{n+1}+F_{n-1}\right)=F_{2 n}\) and \(F_{n}^{2}-F_{n-1} F_{n+1}=(-1)^{n+1}\). Thus at the 8th step, the cotangent of the angle is \(F_{16}=987\).
|
987
|
HMMT_2
|
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5
|
Suppose \(\triangle A B C\) has lengths \(A B=5, B C=8\), and \(C A=7\), and let \(\omega\) be the circumcircle of \(\triangle A B C\). Let \(X\) be the second intersection of the external angle bisector of \(\angle B\) with \(\omega\), and let \(Y\) be the foot of the perpendicular from \(X\) to \(B C\). Find the length of \(Y C\).
|
Extend ray \(\overrightarrow{A B}\) to a point \(D\), since \(B X\) is an angle bisector, we have \(\angle X B C=\angle X B D=180^{\circ}-\angle X B A=\angle X C A\), so \(X C=X A\) by the inscribed angle theorem. Now, construct a point \(E\) on \(B C\) so that \(C E=A B\). Since \(\angle B A X \cong \angle B C X\), we have \(\triangle B A X \cong \triangle E C X\) by SAS congruence. Thus, \(X B=X E\), so \(Y\) bisects segment \(B E\). Since \(B E=B C-E C=8-5=3\), we have \(Y C=E C+Y E=5+\frac{1}{2} \cdot 3=\frac{13}{2}\). (Archimedes Broken Chord Theorem).
|
\frac{13}{2}
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
A polynomial $P$ has four roots, $\frac{1}{4}, \frac{1}{2}, 2,4$. The product of the roots is 1, and $P(1)=1$. Find $P(0)$.
|
A polynomial $Q$ with $n$ roots, $x_{1}, \ldots, x_{n}$, and $Q\left(x_{0}\right)=1$ is given by $Q(x)=\frac{\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)}{\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right) \cdots\left(x_{0}-x_{4}\right)}$, so $P(0)=\frac{1}{\frac{3}{4} \cdot \frac{1}{2} \cdot(-1) \cdot(-3)}=\frac{8}{9}$.
|
\frac{8}{9}
|
HMMT_2
|
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 5
|
In a wooden block shaped like a cube, all the vertices and edge midpoints are marked. The cube is cut along all possible planes that pass through at least four marked points. Let \(N\) be the number of pieces the cube is cut into. Estimate \(N\). An estimate of \(E>0\) earns \(\lfloor 20 \min (N / E, E / N)\rfloor\) points.
|
Answer: 15600
|
15600
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics",
"Mathematics -> Algebra -> Intermediate Algebra -> Inequalities"
] | 5
|
Find the largest positive integer \(n\) for which there exist \(n\) finite sets \(X_{1}, X_{2}, \ldots, X_{n}\) with the property that for every \(1 \leq a<b<c \leq n\), the equation \(\left|X_{a} \cup X_{b} \cup X_{c}\right|=\lceil\sqrt{a b c}\rceil\) holds.
|
First, we construct an example for \(N=4\). Let \(X_{1}, X_{2}, X_{3}, X_{4}\) be pairwise disjoint sets such that \(X_{1}=\varnothing,\left|X_{2}\right|=1,\left|X_{3}\right|=2\), and \(\left|X_{4}\right|=2\). It is straightforward to verify the condition. We claim that there are no five sets \(X_{1}, X_{2}, \ldots, X_{5}\) for which \(#\left(X_{a} \cup X_{b} \cup X_{c}\right)=\lceil\sqrt{a b c}\rceil\), for \(1 \leq a<b<c \leq 5\). Note that showing the non-existence of five such sets implies that there are no \(n\) sets with the desired property for \(n \geq 5\) as well. Suppose, for sake of contradiction, that there are such \(X_{1}, \ldots, X_{5}\). Then, note that \(\left|X_{1} \cup X_{2} \cup X_{4}\right|=3\), \(\left|X_{1} \cup X_{2} \cup X_{5}\right|=4\), and \(\left|X_{2} \cup X_{4} \cup X_{5}\right|=7\). Note that \(\left|X_{1} \cup X_{2} \cup X_{4}\right|+\left|X_{1} \cup X_{2} \cup X_{5}\right|=\left|X_{2} \cup X_{4} \cup X_{5}\right|\). For any sets \(A, B, C, D\), we have the following two inequalities: \(|A \cup B \cup C|+|A \cup B \cup D| \geq|A \cup B \cup C \cup D| \geq|B \cup C \cup D|\). For \(A=X_{1}, B=X_{2}, C=X_{4}\), and \(D=X_{5}\) in the situation above, we conclude that the equalities must both hold in both inequalities. The first equality shows that \(X_{1} \cup X_{2}=\varnothing\), and therefore both \(X_{1}\) and \(X_{2}\) are empty. Now observe that \(\left|X_{1} \cup X_{4} \cup X_{5}\right|=5 \neq 7=\left|X_{2} \cup X_{4} \cup X_{5}\right|\). This gives a contradiction.
|
4
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Suppose there are 100 cookies arranged in a circle, and 53 of them are chocolate chip, with the remainder being oatmeal. Pearl wants to choose a contiguous subsegment of exactly 67 cookies and wants this subsegment to have exactly \(k\) chocolate chip cookies. Find the sum of the \(k\) for which Pearl is guaranteed to succeed regardless of how the cookies are arranged.
|
We claim that the only values of \(k\) are 35 and 36. WLOG assume that the cookies are labelled 0 through 99 around the circle. Consider the following arrangement: cookies 0 through 17,34 through 50, and 67 through 84 are chocolate chip, and the remaining are oatmeal. (The cookies form six alternating blocks around the circle of length \(18,16,17,16,18,15\).) Consider the block of 33 cookies that are not chosen. It is not difficult to see that since the sum of the lengths of each two adjacent block is always at least 33 and at most 34, this block of unchosen cookies always contains at least one complete block of cookies of the same type (and no other cookies of this type). So this block contains 17 or 18 or \(33-16=17\) or \(33-15=18\) chocolate chip cookies. Therefore, the block of 67 chosen cookies can only have \(53-17=36\) or \(53-18=35\) chocolate chip cookies. Now we show that 35 and 36 can always be obtained. Consider all possible ways to choose 67 cookies: cookies 0 through 66,1 through \(67, \ldots, 99\) through 65. It is not difficult to see that the number of chocolate chip cookies in the block changes by at most 1 as we advance from one way to the next. Moreover, each cookie will be chosen 67 times, so on average there will be \(\frac{67.53}{100}=35.51\) chocolate chip cookies in each block. Since not all blocks are below average and not all blocks are above average, there must be a point where a block below average transition into a block above average. The difference of these two blocks is at most 1, so one must be 35 and one must be 36. Therefore, the sum of all possible values of \(k\) is \(35+36=71\).
|
71
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5
|
Five people take a true-or-false test with five questions. Each person randomly guesses on every question. Given that, for each question, a majority of test-takers answered it correctly, let $p$ be the probability that every person answers exactly three questions correctly. Suppose that $p=\frac{a}{2^{b}}$ where $a$ is an odd positive integer and $b$ is a nonnegative integer. Compute 100a+b.
|
There are a total of $16^{5}$ ways for the people to collectively ace the test. Consider groups of people who share the same problems that they got incorrect. We either have a group of 2 and a group of 3 , or a group 5 . In the first case, we can pick the group of two in $\binom{5}{2}$ ways, the problems they got wrong in $\binom{5}{2}$ ways. Then there are 3! ways for the problems of group 3. There are 600 cases here. In the second case, we can $5!\cdot 4!/ 2=120 \cdot 12$ ways to organize the five cycle ( $4!/ 2$ to pick a cycle and 5 ! ways to assign a problem to each edge in the cycle). Thus, the solution is $\frac{255}{2^{17}}$ and the answer is 25517.
|
25517
|
HMMT_2
|
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes",
"Mathematics -> Precalculus -> Trigonometric Functions"
] | 5
|
Let $P A B C$ be a tetrahedron such that $\angle A P B=\angle A P C=\angle B P C=90^{\circ}, \angle A B C=30^{\circ}$, and $A P^{2}$ equals the area of triangle $A B C$. Compute $\tan \angle A C B$.
|
Observe that $$\begin{aligned} \frac{1}{2} \cdot A B \cdot A C \cdot \sin \angle B A C & =[A B C]=A P^{2} \\ & =\frac{1}{2}\left(A B^{2}+A C^{2}-B C^{2}\right) \\ & =A B \cdot A C \cdot \cos \angle B A C \end{aligned}$$ so $\tan \angle B A C=2$. Also, we have $\tan \angle A B C=\frac{1}{\sqrt{3}}$. Also, for any angles $\alpha, \beta, \gamma$ summing to $180^{\circ}$, one can see that $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \cdot \tan \beta \cdot \tan \gamma$. Thus we have $\tan \angle A C B+2+\frac{1}{\sqrt{3}}=$ $\tan \angle A C B \cdot 2 \cdot \frac{1}{\sqrt{3}}$, so $\tan \angle A C B=8+5 \sqrt{3}$.
|
8+5 \sqrt{3}
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Mathematical Statistics"
] | 5
|
679 contestants participated in HMMT February 2017. Let \(N\) be the number of these contestants who performed at or above the median score in at least one of the three individual tests. Estimate \(N\). An estimate of \(E\) earns \(\left\lfloor 20-\frac{|E-N|}{2}\right\rfloor\) or 0 points, whichever is greater.
|
Out of the 679 total contestants at HMMT February 2017, 188 contestants scored at least the median on all three tests, 159 contestants scored at least the median on two tests, and 169 contestants scored at least the median on one test, giving a total of 516 contestants
|
516
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Other",
"Mathematics -> Algebra -> Intermediate Algebra -> Logarithmic Functions"
] | 5
|
How many positive integers $n \leq 2009$ have the property that $\left\lfloor\log _{2}(n)\right\rfloor$ is odd?
|
We wish to find $n$ such that there is some natural number $k$ for which $2 k-1 \leq \log _{2} n<$ $2 k$. Since $n \leq 2009$ we must have $k \leq 5$. This is equivalent to finding the number of positive integers $n \leq 2009$ satisfying $2^{2 k-1} \leq n<2^{2 k}$ for some $k \leq 5$, so the number of such integers is $2+2^{3}+2^{5}+2^{7}+2^{9}=682$
|
682
|
HMMT_2
|
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 5
|
Let $A X B Y$ be a cyclic quadrilateral, and let line $A B$ and line $X Y$ intersect at $C$. Suppose $A X \cdot A Y=6, B X \cdot B Y=5$, and $C X \cdot C Y=4$. Compute $A B^{2}$.
|
Observe that $$\begin{aligned} & \triangle A C X \sim \triangle Y C B \Longrightarrow \frac{A C}{A X}=\frac{C Y}{B Y} \\ & \triangle A C Y \sim \triangle X C B \Longrightarrow \frac{A C}{A Y}=\frac{C X}{B X} \end{aligned}$$ Mulitplying these two equations together, we get that $$A C^{2}=\frac{(C X \cdot C Y)(A X \cdot A Y)}{B X \cdot B Y}=\frac{24}{5}$$ Analogously, we obtain that $$B C^{2}=\frac{(C X \cdot C Y)(B X \cdot B Y)}{A X \cdot A Y}=\frac{10}{3}$$ Hence, we have $$A B=A C+B C=\sqrt{\frac{24}{5}}+\sqrt{\frac{10}{3}}=\frac{11 \sqrt{30}}{15}$$ implying the answer.
|
\frac{242}{15}
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other"
] | 5
|
There are 2018 frogs in a pool and there is 1 frog on the shore. In each time-step thereafter, one random frog moves position. If it was in the pool, it jumps to the shore, and vice versa. Find the expected number of time-steps before all frogs are in the pool for the first time.
|
Consider the general case of \(n\) frogs. Let \(E_{i}\) be the expected time for all frogs to enter the pool when \(i\) frogs are on the shore and \(n-i\) frogs are in the pool. We have \(E_{0}=0, E_{n}=1+E_{n-1}\), and \(E_{i}=\frac{i}{n} E_{i-1}+\frac{n-i}{n} E_{i+1}+1\) for \(0<i<n\). Define \(f_{i}\) so that \(E_{i}=\frac{f_{i}}{(n-1)(n-2) \cdots(i)}+E_{i-1}\). Then by plugging this equation into the first equation, we can show that \(f_{i}=n(n-1) \cdots(i+1)+(n-i) f_{i+1}\). Furthermore, we know that \(f_{n}=1\). Therefore \(f_{1} =\sum_{i=1}^{n} \frac{n!}{i!} \frac{(n-1)!}{(n-i)!} =(n-1)!\sum_{i=1}^{n}\binom{n}{i} =(n-1)!\left(2^{n}-1\right)\). Therefore \(E_{1}=\frac{(n-1)!\left(2^{n}-1\right)}{(n-1)!}+E_{0}=2^{n}-1\). Plugging in \(n=2018\) yields \(E_{1}=2^{2018}-1\).
|
2^{2018}-1
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5
|
Randall proposes a new temperature system called Felsius temperature with the following conversion between Felsius \(^{\circ} \mathrm{E}\), Celsius \(^{\circ} \mathrm{C}\), and Fahrenheit \(^{\circ} \mathrm{F}\): \(^{\circ} E=\frac{7 \times{ }^{\circ} \mathrm{C}}{5}+16=\frac{7 \times{ }^{\circ} \mathrm{F}-80}{9}\). For example, \(0^{\circ} \mathrm{C}=16^{\circ} \mathrm{E}\). Let \(x, y, z\) be real numbers such that \(x^{\circ} \mathrm{C}=x^{\circ} \mathrm{E}, y^{\circ} E=y^{\circ} \mathrm{F}, z^{\circ} \mathrm{C}=z^{\circ} F\). Find \(x+y+z\).
|
Notice that \((5 k)^{\circ} \mathrm{C}=(7 k+16)^{\circ} E=(9 k+32)^{\circ} \mathrm{F}\), so Felsius is an exact average of Celsius and Fahrenheit at the same temperature. Therefore we conclude that \(x=y=z\), and it is not difficult to compute that they are all equal to -40.
|
-120
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Sequences and Series"
] | 5
|
Let \{a_{n}\}_{n \geq 1}$ be an arithmetic sequence and \{g_{n}\}_{n \geq 1}$ be a geometric sequence such that the first four terms of \{a_{n}+g_{n}\}$ are $0,0,1$, and 0 , in that order. What is the 10th term of \{a_{n}+g_{n}\}$ ?
|
Let the terms of the geometric sequence be $a, r a, r^{2} a, r^{3} a$. Then, the terms of the arithmetic sequence are $-a,-r a,-r^{2} a+1,-r^{3} a$. However, if the first two terms of this sequence are $-a,-r a$, the next two terms must also be $(-2 r+1) a,(-3 r+2) a$. It is clear that $a \neq 0$ because $a_{3}+g_{3} \neq 0$, so $-r^{3}=-3 r+2 \Rightarrow r=1$ or -2 . However, we see from the arithmetic sequence that $r=1$ is impossible, so $r=-2$. Finally, by considering $a_{3}$, we see that $-4 a+1=5 a$, so $a=1 / 9$. We also see that $a_{n}=(3 n-4) a$ and $g_{n}=(-2)^{n-1} a$, so our answer is $a_{10}+g_{10}=(26-512) a=-486 a=-54$.
|
-54
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5
|
Let \(a, b, c\) be positive integers. All the roots of each of the quadratics \(a x^{2}+b x+c, a x^{2}+b x-c, a x^{2}-b x+c, a x^{2}-b x-c\) are integers. Over all triples \((a, b, c)\), find the triple with the third smallest value of \(a+b+c\).
|
The quadratic formula yields that the answers to these four quadratics are \(\frac{ \pm b \pm \sqrt{b^{2} \pm 4 a c}}{2 a}\). Given that all eight of these expressions are integers, we can add or subtract appropriate pairs to get that \(\frac{b}{a}\) and \(\frac{\sqrt{b^{2} \pm 4 a c}}{a}\) are integers. Let \(b^{\prime}=\frac{b}{a}\) and \(c^{\prime}=\frac{4 c}{a}\). We can rewrite the expressions to get that \(b^{\prime}\) and \(\sqrt{b^{\prime 2} \pm c^{\prime}}\) are positive integers, which also tells us that \(c^{\prime}\) is a positive integer. Let \(b^{\prime 2}+c^{\prime}=n^{2}\), \(b^{\prime 2}-c^{\prime}=m^{2}\). Notice that \(a+b+c=a\left(1+b^{\prime}+\frac{c^{\prime}}{4}\right)\), so to find the third smallest value of \(a+b+c\), we first find small solutions to \(\left(b^{\prime}, c^{\prime}\right)\). To do this, we find triples \(\left(m, b^{\prime}, n\right)\) such that \(m^{2}, b^{\prime 2}, n^{2}\) form an arithmetic sequence. Because odd squares are \(1 \bmod 4\) and even squares are \(0 \bmod 4\), if any of these three terms is odd, then all three terms must be odd. By dividing these terms by the largest possible power of 2 then applying the same logic, we can extend our result to conclude that \(v_{2}(m)=v_{2}\left(b^{\prime}\right)=v_{2}(n)\). Thus, we only need to look at \(\left(m, b^{\prime}, n\right)\) all odd, then multiply them by powers of 2 to get even solutions. We then plug in \(b^{\prime}=3,5,7,9\), and find that out of these options, only \(\left(n, b^{\prime}, m\right)=(1,5,7)\) works, giving \(\left(b^{\prime}, c^{\prime}\right)=(5,24), a+b+c=12 a\). Multiplying by 2 yields that \(\left(n, b^{\prime}, m\right)=(2,10,14)\) also works, giving \(\left(b^{\prime}, c^{\prime}\right)=(10,96), a+b+c=35 a\). For \(11 \leq b \leq 17\), we can check that \(m=b+2\) fails to give an integer n. For \(11 \leq b \leq 17, m \neq b+2, a+b+c=a\left(1+b^{\prime}+\frac{c^{\prime}}{4}\right) \geq a\left(1+11+\frac{15^{2}-11^{2}}{4}\right)=38 a\), the smallest possible value of which is greater than \(12 a\) with \(a=1,12 a\) with \(a=2\), and \(35 a\) with \(a=1\). Thus, it cannot correspond to the solution with the third smallest \(a+b+c\). For \(b \geq 19\), \(a+b+c=a\left(1+b^{\prime}+\frac{c^{\prime}}{4}\right) \geq a\left(1+19+\frac{21^{2}+19^{2}}{4}\right)=40 a\), which, similar as before, can't correspond to the solution with the third smallest \(a+b+c\). Thus the smallest solution is \(\left(a, b^{\prime}, c^{\prime}\right)=(1,5,24),(a, b, c)=(1,5,6)\), the second smallest solution is \(\left(a, b^{\prime}, c^{\prime}\right)=(2,5,24),(a, b, c)=(2,10,12)\), and the third smallest solution that the problem asks for is \(\left(a, b^{\prime}, c^{\prime}\right)=(1,10,96),(a, b, c)=(1,10,24)\).
|
(1,10,24)
|
HMMT_2
|
[
"Mathematics -> Precalculus -> Trigonometric Functions",
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers"
] | 5
|
Let $\alpha, \beta$, and $\gamma$ be three real numbers. Suppose that $\cos \alpha+\cos \beta+\cos \gamma =1$ and $\sin \alpha+\sin \beta+\sin \gamma =1$. Find the smallest possible value of $\cos \alpha$.
|
Let $a=\cos \alpha+i \sin \alpha, b=\cos \beta+i \sin \beta$, and $c=\cos \gamma+i \sin \gamma$. We then have $a+b+c=1+i$ where $a, b, c$ are complex numbers on the unit circle. Now, to minimize $\cos \alpha=\operatorname{Re}[a]$, consider a triangle with vertices $a, 1+i$, and the origin. We want $a$ as far away from $1+i$ as possible while maintaining a nonnegative imaginary part. This is achieved when $b$ and $c$ have the same argument, so $|b+c|=|1+i-a|=2$. Now $a, 0$, and $1+i$ form a $1-2-\sqrt{2}$ triangle. The value of $\cos \alpha$ is now the cosine of the angle between the 1 and $\sqrt{2}$ sides plus the $\frac{\pi}{4}$ angle from $1+i$. Call the first angle $\delta$. Then $\cos \delta =\frac{1^{2}+(\sqrt{2})^{2}-2^{2}}{2 \cdot 1 \cdot \sqrt{2}} =\frac{-1}{2 \sqrt{2}}$ and $\cos \alpha =\cos \left(\frac{\pi}{4}+\delta\right) =\cos \frac{\pi}{4} \cos \delta-\sin \frac{\pi}{4} \sin \delta =\frac{\sqrt{2}}{2} \cdot \frac{-1}{2 \sqrt{2}}-\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{7}}{2 \sqrt{2}} =\frac{-1-\sqrt{7}}{4}$.
|
\frac{-1-\sqrt{7}}{4}
|
HMMT_2
|
[
"Mathematics -> Algebra -> Intermediate Algebra -> Other",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Let $S$ be the set of integers of the form $2^{x}+2^{y}+2^{z}$, where $x, y, z$ are pairwise distinct non-negative integers. Determine the 100th smallest element of $S$.
|
S is the set of positive integers with exactly three ones in its binary representation. The number of such integers with at most $d$ total bits is \binom{d}{3}$, and noting that \binom{9}{3}=84$ and \binom{10}{3}=120$, we want the 16th smallest integer of the form $2^{9}+2^{x}+2^{y}$, where $y<x<9$. Ignoring the $2^{9}$ term, there are \binom{d^{\prime}}{2}$ positive integers of the form $2^{x}+2^{y}$ with at most $d^{\prime}$ total bits. Because \binom{6}{2}=15$, our answer is $2^{9}+2^{6}+2^{0}=577$. (By a bit, we mean a digit in base 2 .)
|
577
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Let $S_{0}$ be a unit square in the Cartesian plane with horizontal and vertical sides. For any $n>0$, the shape $S_{n}$ is formed by adjoining 9 copies of $S_{n-1}$ in a $3 \times 3$ grid, and then removing the center copy. Let $a_{n}$ be the expected value of $\left|x-x^{\prime}\right|+\left|y-y^{\prime}\right|$, where $(x, y)$ and $\left(x^{\prime}, y^{\prime}\right)$ are two points chosen randomly within $S_{n}$. There exist relatively prime positive integers $a$ and $b$ such that $$\lim _{n \rightarrow \infty} \frac{a_{n}}{3^{n}}=\frac{a}{b}$$ Compute $100 a+b$.
|
By symmetry, we only need to consider the $x$-distance, then we can multiply our answer by 2. Let this quantity be $g(n)=a_{n} / 2$. Divide the $n$th iteration fractal into three meta-columns of equal width. Then the probability that a random point is in the first, second, and third meta-columns is $\frac{3}{8}, \frac{2}{8}$, and $\frac{3}{8}$, respectively. If the two points end up in neighboring meta columns, the expected value of their $x$-distance is simply the width of a meta-column, which is $3^{n-1}$. If they end up in opposite meta-columns (the left and right ones), it is twice this amount, which is $2 \cdot 3^{n-1}$. Finally, if the two points lie in the same meta-column, which happens with probability $\left(\frac{3}{8}\right)^{2}+\left(\frac{2}{8}\right)^{2}+\left(\frac{3}{8}\right)^{2}=\frac{11}{32}$, the expected $x$-distance is just $g(n-1)$. Thus, we have $$g(n)=3^{n-1}\left(2 \cdot \frac{3}{8} \cdot \frac{2}{8}+2 \cdot \frac{2}{8} \cdot \frac{3}{8}\right)+\left(2 \cdot 3^{n-1}\right)\left(2 \cdot \frac{3}{8} \cdot \frac{3}{8}\right)+\frac{11}{32} g(n-1)=\frac{15}{16} \cdot 3^{n-1}+\frac{11}{32} g(n-1)$$ As $n$ grows, say this is asymptotic to $g(n)=3^{n} C$. For some constant $C$. Then we can write $3^{n} C=\frac{15}{16} \cdot 3^{n-1}+\frac{11}{32} \cdot 3^{n-1} C \Longrightarrow C=\frac{6}{17}$. Our final answer is twice this, which is $\frac{12}{17}$.
|
1217
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other",
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5
|
Rachel has the number 1000 in her hands. When she puts the number $x$ in her left pocket, the number changes to $x+1$. When she puts the number $x$ in her right pocket, the number changes to $x^{-1}$. Each minute, she flips a fair coin. If it lands heads, she puts the number into her left pocket, and if it lands tails, she puts it into her right pocket. She then takes the new number out of her pocket. If the expected value of the number in Rachel's hands after eight minutes is $E$, then compute $\left\lfloor\frac{E}{10}\right\rfloor$.
|
Call a real number very large if $x \in[1000,1008]$, very small if $x \in\left[0, \frac{1}{1000}\right]$, and medium-sized if $x \in\left[\frac{1}{8}, 8\right]$. Every number Rachel is ever holding after at most 8 steps will fall under one of these categories. Therefore the main contribution to $E$ will come from the probability that Rachel is holding a number at least 1000 at the end. Note that if her number ever becomes medium-sized, it will never become very large or very small again. Therefore the only way her number ends up above 1000 is if the sequence of moves consists of $x \rightarrow x+1$ moves and consecutive pairs of $x \rightarrow x^{-1}$ moves. Out of the 256 possible move sequences, the number of ways for the number to stay above 1000 is the number of ways of partitioning 8 into an ordered sum of 1 and 2, or the ninth Fibonacci number $F_{9}=34$. Therefore $$\frac{34}{256} \cdot 1000 \leq E \leq \frac{34}{256} \cdot 1000+8$$ where $\frac{34}{256} \cdot 1000 \approx 132.8$. Furthermore, the extra contribution will certainly not exceed 7 , so we get that $\left\lfloor\frac{E}{10}\right\rfloor=13$.
|
13
|
HMMT_2
|
[
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers"
] | 5
|
Let $x, y$ be complex numbers such that \frac{x^{2}+y^{2}}{x+y}=4$ and \frac{x^{4}+y^{4}}{x^{3}+y^{3}}=2$. Find all possible values of \frac{x^{6}+y^{6}}{x^{5}+y^{5}}$.
|
Let $A=\frac{1}{x}+\frac{1}{y}$ and let $B=\frac{x}{y}+\frac{y}{x}$. Then $$ \frac{B}{A}=\frac{x^{2}+y^{2}}{x+y}=4 $$ so $B=4 A$. Next, note that $$ B^{2}-2=\frac{x^{4}+y^{4}}{x^{2} y^{2}} \text { and } A B-A=\frac{x^{3}+y^{3}}{x^{2} y^{2}} $$ so $$ \frac{B^{2}-2}{A B-A}=2 $$ Substituting $B=4 A$ and simplifying, we find that $4 A^{2}+A-1=0$, so $A=\frac{-1 \pm \sqrt{17}}{8}$. Finally, note that $$ 64 A^{3}-12 A=B^{3}-3 B=\frac{x^{6}+y^{6}}{x^{3} y^{3}} \text { and } 16 A^{3}-4 A^{2}-A=A\left(B^{2}-2\right)-(A B-A)=\frac{x^{5}+y^{5}}{x^{3} y^{3}} $$ $$ \frac{x^{6}+y^{6}}{x^{5}+y^{5}}=\frac{64 A^{2}-12}{16 A^{2}-4 A-1}=\frac{4-16 A}{3-8 A} $$ where the last inequality follows from the fact that $4 A^{2}=1-A$. If $A=\frac{-1+\sqrt{17}}{8}$, then this value equals $10+2 \sqrt{17}$. Similarly, if $A=\frac{-1-\sqrt{17}}{8}$, then this value equals $10-2 \sqrt{17}$. (It is not hard to see that these values are achievable by noting that with the values of $A$ and $B$ we can solve for $x+y$ and $x y$, and thus for $x$ and $y$.)
|
10 \pm 2 \sqrt{17}
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities",
"Mathematics -> Algebra -> Intermediate Algebra -> Quadratic Functions"
] | 5
|
Let $x, y$, and $N$ be real numbers, with $y$ nonzero, such that the sets $\left\{(x+y)^{2},(x-y)^{2}, x y, x / y\right\}$ and $\{4,12.8,28.8, N\}$ are equal. Compute the sum of the possible values of $N$.
|
First, suppose that $x$ and $y$ were of different signs. Then $x y<0$ and $x / y<0$, but the set has at most one negative value, a contradiction. Hence, $x$ and $y$ have the same sign; without loss of generality, we say $x$ and $y$ are both positive. Let $(s, d):=(x+y, x-y)$. Then the set given is equal to $\left\{s^{2}, d^{2}, \frac{1}{4}\left(s^{2}-d^{2}\right), \frac{s+d}{s-d}\right\}$. We split into two cases: - Case 1: $\frac{s+d}{s-d}=N$. This forces $s^{2}=28.8$ and $d^{2}=12.8$, since $\frac{1}{4}(28.8-12.8)=4$. Then $s=12 \sqrt{0.2}$ and $d= \pm 8 \sqrt{0.2}$, so $N$ is either $\frac{12+8}{12-8}=5$ or $\frac{12-8}{12+8}=0.2$. - Case 2: $\frac{s+d}{s-d} \neq N$. Suppose $\frac{s+d}{s-d}=k$, so $(s, d)=((k+1) t,(k-1) t)$ for some $t$. Then $s^{2}: d^{2}$ : $\frac{1}{4}\left(s^{2}-d^{2}\right)=(k+1)^{2}:(k-1)^{2}: k$. Trying $k=4,12.8,28.8$ reveals that only $k=4$ is possible, since $28.8: 12.8=(4-1)^{2}: 4$. This forces $N=s^{2}=\frac{5^{2}}{4} \cdot 12.8=80$. Hence, our final total is $5+0.2+80=85.2$
|
85.2
|
HMMT_2
|
[
"Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions"
] | 5
|
A sequence $\left\{a_{n}\right\}_{n \geq 1}$ of positive reals is defined by the rule $a_{n+1} a_{n-1}^{5}=a_{n}^{4} a_{n-2}^{2}$ for integers $n>2$ together with the initial values $a_{1}=8$ and $a_{2}=64$ and $a_{3}=1024$. Compute $$\sqrt{a_{1}+\sqrt{a_{2}+\sqrt{a_{3}+\cdots}}}$$
|
Taking the base-2 $\log$ of the sequence $\left\{a_{n}\right\}$ converts the multiplicative rule to a more familiar additive rule: $\log _{2}\left(a_{n+1}\right)-4 \log _{2}\left(a_{n}\right)+5 \log _{2}\left(a_{n-1}\right)-2 \log _{2}\left(a_{n-2}\right)=0$. The characteristic equation is $0=x^{3}-4 x^{2}+5 x-2=(x-1)^{2}(x-2)$, so $\log _{2}\left(a_{n}\right)$ is of the form $a \cdot n+b+c \cdot 2^{n}$ and we find $a_{n}=2^{2 n+2^{n-1}}$. Now, $$\sqrt{a_{1}+\sqrt{a_{2}+\sqrt{a_{3}+\cdots}}}=\sqrt{2} \cdot \sqrt{4+\sqrt{16+\sqrt{64+\cdots}}}$$ We can estimate the new nested radical expression as 3, which expands thus $$3=\sqrt{4+5}=\sqrt{4+\sqrt{16+9}}=\sqrt{4+\sqrt{16+\sqrt{64+17}}}=\cdots$$ As a rigorous confirmation, we have $2^{k}+1=\sqrt{4^{k}+\left(2^{k+1}+1\right)}$, as desired. It follows that the answer is $3 \sqrt{2}$.
|
3\sqrt{2}
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
Let $x$ and $y$ be positive real numbers. Define $a=1+\frac{x}{y}$ and $b=1+\frac{y}{x}$. If $a^{2}+b^{2}=15$, compute $a^{3}+b^{3}$.
|
Note that $a-1=\frac{x}{y}$ and $b-1=\frac{y}{x}$ are reciprocals. That is, $$(a-1)(b-1)=1 \Longrightarrow a b-a-b+1=1 \Longrightarrow a b=a+b$$ Let $t=a b=a+b$. Then we can write $$a^{2}+b^{2}=(a+b)^{2}-2 a b=t^{2}-2 t$$ so $t^{2}-2 t=15$, which factors as $(t-5)(t+3)=0$. Since $a, b>0$, we must have $t=5$. Then, we compute $$a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)=5^{3}-3 \cdot 5^{2}=50$$
|
50
|
HMMT_2
|
[
"Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions"
] | 5
|
Suppose $a, b, c$, and $d$ are pairwise distinct positive perfect squares such that $a^{b}=c^{d}$. Compute the smallest possible value of $a+b+c+d$.
|
Note that if $a$ and $c$ are divisible by more than one distinct prime, then we can just take the prime powers of a specific prime. Thus, assume $a$ and $c$ are powers of a prime $p$. Assume $a=4^{x}$ and $c=4^{y}$. Then $x b=y d$. Because $b$ and $d$ are squares, the ratio of $x$ to $y$ is a square, so assume $x=1$ and $y=4$. We can't take $b=4$ and $c=1$, but we instead can take $b=36$ and $c=9$. It can be checked that other values of $x$ and $y$ are too big. This gives $4^{36}=256^{9}$, which gives a sum of 305. If $a$ and $c$ are powers of 9 , then $\max (a, c) \geq 9^{4}$, which is already too big. Thus, 305 is optimal.
|
305
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5
|
Fran writes the numbers \(1,2,3, \ldots, 20\) on a chalkboard. Then she erases all the numbers by making a series of moves; in each move, she chooses a number \(n\) uniformly at random from the set of all numbers still on the chalkboard, and then erases all of the divisors of \(n\) that are still on the chalkboard (including \(n\) itself). What is the expected number of moves that Fran must make to erase all the numbers?
|
For each \(n, 1 \leq n \leq 20\), consider the first time that Fran chooses one of the multiples of \(n\). It is in this move that \(n\) is erased, and all the multiples of \(n\) at most 20 are equally likely to be chosen for this move. Hence this is the only move in which Fran could possibly choose \(n\); since there are \(\lfloor 20 / n\rfloor\) multiples of \(n\) at most 20, this means that the probability that \(n\) is ever chosen is \(1 /\lfloor 20 / n\rfloor\). Therefore the expected number of moves is \(E =\sum_{n=1}^{20} \frac{1}{\lfloor 20 / n\rfloor} =\frac{1}{20}+\frac{1}{10}+\frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+4\left(\frac{1}{2}\right)+10(1)=\frac{131}{10}\). (This sum is easier to compute than it may seem, if one notes that \(1 / 20+1 / 5+1 / 4=1 / 2\) and \(1 / 6+1 / 3=1 / 2)\)
|
\frac{131}{10}
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Algebra -> Intermediate Algebra -> Other"
] | 5
|
Let $N$ be a positive integer whose decimal representation contains 11235 as a contiguous substring, and let $k$ be a positive integer such that $10^{k}>N$. Find the minimum possible value of $$ \frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)} $$
|
Set $m=\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}$. Then, in lowest terms, $\frac{N}{10^{k}-1}=\frac{a}{m}$ for some integer $a$. On the other hand, the decimal expansion of $\frac{N}{10^{k}-1}$ simply consists of the decimal expansion of $N$, possibly with some padded zeros, repeating. Since $N$ contains 11235 as a contiguous substring, the decimal representation of $\frac{a}{m}$ must as well. Conversely, if $m$ is relatively prime to 10 and if there exists an $a$ such that the decimal representation of $\frac{a}{m}$ contains the substring 11235 , we claim that $m$ is an attainable value for $\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}$. To see this, note that since $m$ is relatively prime to 10 , there exists a value of $k$ such that $m$ divides $10^{k}-1$ (for example, $k=\phi(m)$ ). Letting $m s=10^{k}-1$ and $N=a s$, it follows that $\frac{a}{m}=\frac{a s}{m s}=\frac{N}{10^{k}-1}$. Since the decimal expansion of this fraction contains the substring 11235, it follows that $N$ must also, and therefore $m$ is an attainable value. We are therefore looking for a fraction $\frac{a}{m}$ which contains the substring 11235 in its decimal expansion. Since 1, 1, 2, 3, and 5 are the first five Fibonacci numbers, it makes sense to look at the value of the infinite series $$ \sum_{i=1}^{\infty} \frac{F_{i}}{10^{i}} $$ A simple generating function argument shows that $\sum_{i=1}^{\infty} F_{i} x^{i}=\frac{x}{1-x-x^{2}}$, so substituting $x=1 / 10$ leads us to the fraction 10/89 (which indeed begins $0.11235 \ldots$ ). How do we know no smaller values of $m$ are possible? Well, if $a^{\prime} / m^{\prime}$ contains the substring 11235 somewhere in its infinitely repeating decimal expansion, then note that there is an $i$ such that the decimal expansion of the fractional part of $10^{i}\left(a^{\prime} / m^{\prime}\right)$ begins with $0.11235 \ldots$ We can therefore, without loss of generality, assume that the decimal representation of $a^{\prime} / m^{\prime}$ begins $0.11235 \ldots$ But since the decimal representation of $10 / 89$ begins $0.11235 \ldots$, it follows that $$ \left|\frac{10}{89}-\frac{a^{\prime}}{m^{\prime}}\right| \leq 10^{-5} $$ On the other hand, this absolute difference, if non-zero, is at least $\frac{1}{89 m^{\prime}}$. If $m^{\prime}<89$, this is at least $\frac{1}{89^{2}}>10^{-5}$, and therefore no smaller values of $m^{\prime}$ are possible.
|
89
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5
|
Suppose $a$ and $b$ are positive integers. Isabella and Vidur both fill up an $a \times b$ table. Isabella fills it up with numbers $1,2, \ldots, a b$, putting the numbers $1,2, \ldots, b$ in the first row, $b+1, b+2, \ldots, 2 b$ in the second row, and so on. Vidur fills it up like a multiplication table, putting $i j$ in the cell in row $i$ and column $j$. Isabella sums up the numbers in her grid, and Vidur sums up the numbers in his grid; the difference between these two quantities is 1200. Compute $a+b$.
|
Using the formula $1+2+\cdots+n=\frac{n(n+1)}{2}$, we get $$\begin{aligned} \frac{a b(a b+1)}{2}-\frac{a(a+1)}{2} \cdot \frac{b(b+1)}{2} & =\frac{a b(2(a b+1)-(a+1)(b+1))}{4} \\ & =\frac{a b(a b-a-b+1)}{4} \\ & =\frac{a b(a-1)(b-1)}{4} \\ & =\frac{a(a-1)}{2} \cdot \frac{b(b-1)}{2} \end{aligned}$$ This means we can write the desired equation as $$a(a-1) \cdot b(b-1)=4800$$ Assume $b \leq a$, so we know $b(b-1) \leq a(a-1)$, so $b(b-1)<70$. Thus, $b \leq 8$. If $b=7$ or $b=8$, then $b(b-1)$ has a factor of 7, which 4800 does not, so $b \leq 6$. If $b=6$ then $b(b-1)=30$, so $a(a-1)=160$, which can be seen to have no solutions. If $b=5$ then $b(b-1)=20$, so $a(a-1)=240$, which has the solution $a=16$, giving $5+16=21$. We need not continue since we are guaranteed only one solution, but we check the remaining cases for completeness. If $b=4$ then $a(a-1)=\frac{4800}{12}=400$, which has no solutions. If $b=3$ then $a(a-1)=\frac{4800}{6}=800$ which has no solutions. Finally, if $b=2$ then $a(a-1)=\frac{4800}{2}=2400$, which has no solutions. The factorization of the left side may come as a surprise; here's a way to see it should factor without doing the algebra. If either $a=1$ or $b=1$, then the left side simplifies to 0. As a result, both $a-1$ and $b-1$ should be a factor of the left side.
|
21
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Determine the number of ways to select a sequence of 8 sets $A_{1}, A_{2}, \ldots, A_{8}$, such that each is a subset (possibly empty) of \{1,2\}, and $A_{m}$ contains $A_{n}$ if $m$ divides $n$.
|
Consider an arbitrary $x \in\{1,2\}$, and let us consider the number of ways for $x$ to be in some of the sets so that the constraints are satisfied. We divide into a few cases: - Case: $x \notin A_{1}$. Then $x$ cannot be in any of the sets. So there is one possibility. - Case: $x \in A_{1}$ but $x \notin A_{2}$. Then the only other sets that $x$ could be in are $A_{3}, A_{5}, A_{7}$, and $x$ could be in some collection of them. There are 8 possibilities in this case. - Case: $x \in A_{2}$. Then $x \in A_{1}$ automatically. There are 4 independent choices to be make here: (1) whether $x \in A_{5} ;(2)$ whether $x \in A_{7} ;(3)$ whether $x \in A_{3}$, and if yes, whether $x \in A_{6}$; (4) whether $x \in A_{4}$, and if yes, whether $x \in A_{8}$. There are $2 \times 2 \times 3 \times 3=36$ choices here. Therefore, there are $1+8+36=45$ ways to place $x$ into some of the sets. Since the choices for $x=1$ and $x=2$ are made independently, we see that the total number of possibilities is $45^{2}=2025$.
|
2025
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Let $P_{1}, P_{2}, \ldots, P_{8}$ be 8 distinct points on a circle. Determine the number of possible configurations made by drawing a set of line segments connecting pairs of these 8 points, such that: (1) each $P_{i}$ is the endpoint of at most one segment and (2) two no segments intersect. (The configuration with no edges drawn is allowed.)
|
Let $f(n)$ denote the number of valid configurations when there are $n$ points on the circle. Let $P$ be one of the points. If $P$ is not the end point of an edge, then there are $f(n-1)$ ways to connect the remaining $n-1$ points. If $P$ belongs to an edge that separates the circle so that there are $k$ points on one side and $n-k-2$ points on the other side, then there are $f(k) f(n-k-2)$ ways of finishing the configuration. Thus, $f(n)$ satisfies the recurrence relation $$f(n)=f(n-1)+f(0) f(n-2)+f(1) f(n-3)+f(2) f(n-4)+\cdots+f(n-2) f(0), n \geq 2$$ The initial conditions are $f(0)=f(1)=1$. Using the recursion, we find that $f(2)=2, f(3)=4, f(4)=$ $9, f(5)=21, f(6)=51, f(7)=127, f(8)=323$.
|
323
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Determine the number of 8-tuples of nonnegative integers $\left(a_{1}, a_{2}, a_{3}, a_{4}, b_{1}, b_{2}, b_{3}, b_{4}\right)$ satisfying $0 \leq a_{k} \leq k$, for each $k=1,2,3,4$, and $a_{1}+a_{2}+a_{3}+a_{4}+2 b_{1}+3 b_{2}+4 b_{3}+5 b_{4}=19$.
|
For each $k=1,2,3,4$, note that set of pairs $\left(a_{k}, b_{k}\right)$ with $0 \leq a_{k} \leq k$ maps bijectively to the set of nonnegative integers through the map $\left(a_{k}, b_{k}\right) \mapsto a_{k}+(k+1) b_{k}$, as $a_{k}$ is simply the remainder of $a_{k}+(k+1) b_{k}$ upon division by $k+1$. By letting $x_{k}=a_{k}+(k+1) b_{k}$, we see that the problem is equivalent to finding the number of quadruples of nonnegative integers $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ such that $x_{1}+x_{2}+x_{3}+x_{4}=19$. This is the same as finding the number of quadruples of positive integers $\left(x_{1}+1, x_{2}+1, x_{3}+1, x_{4}+1\right)$ such that $x_{1}+x_{2}+x_{3}+x_{4}=23$. By a standard "dots and bars" argument, we see that the answer is $\binom{22}{3}=1540$.
|
1540
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics",
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 5
|
A bug is on a corner of a cube. A healthy path for the bug is a path along the edges of the cube that starts and ends where the bug is located, uses no edge multiple times, and uses at most two of the edges adjacent to any particular face. Find the number of healthy paths.
|
There are 6 symmetric ways to choose the first two edges on the path. After these are chosen, all subsequent edges are determined, until the starting corner is reached once again.
|
6
|
HMMT_2
|
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 5
|
Assume we are going to sell a bundle item including one unit of Item 1 and one unit of Item 2 at price p_{12}. The total cost of this item is t(c_{1} + c_{2}), where 0 < t < 1. Assume a visiting customer will purchase one piece of this bundle if (S_{1} + S_{2}) >= p_{12}, and if so, your profit is p_{12} - t(c_{1} + c_{2}). Determine the price p_{12} to maximize the expected profit for each visiting customer. Please provide a formula.
|
The price p_{12} that maximizes the expected return is given by: p_{12}^{*} = \begin{cases} \frac{1}{3}(c_{12} + \sqrt{c_{12}^{2} + 6u_{1}u_{2}}), & c_{12} \in [0, \frac{3}{2}u_{1} - u_{2}] \\ \frac{1}{4}(u_{1} + 2u_{2} + 2c_{12}), & c_{12} \in [\frac{3}{2}u_{1} - u_{2}, u_{2} - \frac{1}{2}u_{1}] \\ \frac{1}{3}(u_{1} + u_{2} + 2c_{12}), & c_{12} \in [u_{2} - \frac{1}{2}u_{1}, u_{1} + u_{2}] \end{cases}
|
p_{12}^{*} = \begin{cases} \frac{1}{3}(c_{12} + \sqrt{c_{12}^{2} + 6u_{1}u_{2}}), & c_{12} \in [0, \frac{3}{2}u_{1} - u_{2}] \\ \frac{1}{4}(u_{1} + 2u_{2} + 2c_{12}), & c_{12} \in [\frac{3}{2}u_{1} - u_{2}, u_{2} - \frac{1}{2}u_{1}] \\ \frac{1}{3}(u_{1} + u_{2} + 2c_{12}), & c_{12} \in [u_{2} - \frac{1}{2}u_{1}, u_{1} + u_{2}] \end{cases}
|
alibaba_global_contest
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Compute the number of ways to tile a $3 \times 5$ rectangle with one $1 \times 1$ tile, one $1 \times 2$ tile, one $1 \times 3$ tile, one $1 \times 4$ tile, and one $1 \times 5$ tile. (The tiles can be rotated, and tilings that differ by rotation or reflection are considered distinct.)
|
Our strategy is to first place the $1 \times 5$ and the $1 \times 4$ tiles since their size restricts their location. We have three cases: - Case 1: first row. There are 4 ways to place the $1 \times 4$ tile. There is an empty cell next to the $1 \times 4$ tile, which can either be occupied by the $1 \times 1$ tile or the $1 \times 2$ tile (see diagram). In both cases, there are 2 ways to place the remaining two tiles, so this gives $4 \cdot 2 \cdot 2=16$ ways. - Case 2: middle row. There are 4 ways to place the $1 \times 4$ tile, and the $1 \times 1$ tile must go next to it. There are 2 ways to place the remaining two tiles, so this gives $4 \cdot 2=8$ ways. - Case 3: bottom row. This is the same as Case 1 up to rotation, so there are also 16 ways to place the tiles here. In total, we have $16+8+16=40$ ways to place the tiles.
|
40
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities",
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
Compute the unique ordered pair $(x, y)$ of real numbers satisfying the system of equations $$\frac{x}{\sqrt{x^{2}+y^{2}}}-\frac{1}{x}=7 \text { and } \frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}=4$$
|
Solution 1: Consider vectors $$\binom{x / \sqrt{x^{2}+y^{2}}}{y / \sqrt{x^{2}+y^{2}}} \text { and }\binom{-1 / x}{1 / y}$$ They are orthogonal and add up to $\binom{7}{4}$, which have length $\sqrt{7^{2}+4^{2}}=\sqrt{65}$. The first vector has length 1, so by Pythagorean's theorem, the second vector has length $\sqrt{65-1}=8$, so we have $$\frac{1}{x^{2}}+\frac{1}{y^{2}}=64 \Longrightarrow \sqrt{x^{2}+y^{2}}= \pm 8 x y$$ However, the first equation indicates that $x<0$, while the second equation indicates that $y>0$, so $x y<0$. Thus, $\sqrt{x^{2}+y^{2}}=-8 x y$. Plugging this into both of the starting equations give $$-\frac{1}{8 y}-\frac{1}{x}=7 \text { and }-\frac{1}{8 x}+\frac{1}{y}=4$$ Solving this gives $(x, y)=\left(-\frac{13}{96}, \frac{13}{40}\right)$, which works. Solution 2: Let $x=r \cos \theta$ and $y=r \sin \theta$. Then our equations read $$\begin{aligned} & \cos \theta-\frac{1}{r \cos \theta}=7 \\ & \sin \theta+\frac{1}{r \sin \theta}=4 \end{aligned}$$ Multiplying the first equation by $\cos \theta$ and the second by $\sin \theta$, and then adding the two gives $7 \cos \theta+$ $4 \sin \theta=1$. This means $$4 \sin \theta=1-7 \cos \theta \Longrightarrow 16 \sin ^{2} \theta=1-14 \cos \theta+49 \cos ^{2} \theta \Longrightarrow 65 \cos ^{2} \theta-14 \cos \theta-15=0$$ This factors as $(13 \cos \theta+5)(5 \cos \theta-3)=0$, so $\cos \theta$ is either $\frac{3}{5}$ or $-\frac{5}{13}$. This means either $\cos \theta=\frac{3}{5}$ and $\sin \theta=-\frac{4}{5}$, or $\cos \theta=-\frac{5}{13}$ and $\sin \theta=\frac{12}{13}$. The first case, plugging back in, makes $r$ a negative number, a contradiction, so we take the second case. Then $x=\frac{1}{\cos \theta-7}=-\frac{13}{96}$ and $y=\frac{1}{4-\sin \theta}=\frac{13}{40}$. The answer is $(x, y)=\left(-\frac{13}{96}, \frac{13}{40}\right)$.
|
(-\frac{13}{96}, \frac{13}{40})
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Factorization"
] | 5
|
Compute the largest positive integer such that $\frac{2007!}{2007^{n}}$ is an integer.
|
Note that $2007=3^{2} \cdot 223$. Using the fact that the number of times a prime $p$ divides $n!$ is given by $$\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^{2}}\right\rfloor+\left\lfloor\frac{n}{p^{3}}\right\rfloor+\cdots$$ it follows that the answer is 9.
|
9
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
There are 2017 jars in a row on a table, initially empty. Each day, a nice man picks ten consecutive jars and deposits one coin in each of the ten jars. Later, Kelvin the Frog comes back to see that $N$ of the jars all contain the same positive integer number of coins (i.e. there is an integer $d>0$ such that $N$ of the jars have exactly $d$ coins). What is the maximum possible value of $N$?
|
Label the jars $1,2, \ldots, 2017$. I claim that the answer is 2014. To show this, we need both a construction and an upper bound. For the construction, for $1 \leq i \leq 201$, put a coin in the jars $10 i+1,10 i+2, \ldots, 10 i+10$. After this, each of the jars $1,2, \ldots, 2010$ has exactly one coin. Now, put a coin in each of the jars $2008,2009, \ldots, 2017$. Now, the jars $1,2, \ldots, 2007,2011,2012, \ldots, 2017$ all have exactly one coin. This gives a construction for $N=2014$ (where $d=1$). Now, we show that this is optimal. Let $c_{1}, c_{2}, \ldots, c_{2017}$ denote the number of coins in each of the jars. For $1 \leq j \leq 10$, define $$s_{j}=c_{j}+c_{j+10}+c_{j+20}+\ldots$$ Note that throughout the process, $s_{1}=s_{2}=\cdots=s_{j}$. It is also easy to check that the sums $s_{1}, s_{2}, \ldots, s_{7}$ each involve 202 jars, while the sums $s_{8}, s_{9}, s_{10}$ each involve 201 jars. Call a jar good if it has exactly $d$ coins. If there are at least 2015 good jars, then one can check that it is forced that at least one of $s_{1}, s_{2}, \ldots, s_{7}$ only involves good jars, and similarly, at least one of $s_{8}, s_{9}, s_{10}$ only involves good jars. But this would mean that $202 d=201 d$ as all $s_{i}$ are equal, contradiction.
|
2014
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5
|
Assume the quartic $x^{4}-a x^{3}+b x^{2}-a x+d=0$ has four real roots $\frac{1}{2} \leq x_{1}, x_{2}, x_{3}, x_{4} \leq 2$. Find the maximum possible value of $\frac{\left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right) x_{4}}{\left(x_{4}+x_{2}\right)\left(x_{4}+x_{3}\right) x_{1}}$ (over all valid choices of $\left.a, b, d\right)$.
|
We can rewrite the expression as $$\frac{x_{4}^{2}}{x_{1}^{2}} \cdot \frac{\left(x_{1}+x_{1}\right)\left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)\left(x_{1}+x_{4}\right)}{\left(x_{4}+x_{1}\right)\left(x_{4}+x_{2}\right)\left(x_{4}+x_{3}\right)\left(x_{4}+x_{4}\right)} \frac{x_{4}^{2}}{x_{1}^{2}} \cdot \frac{f\left(-x_{1}\right)}{f\left(-x_{4}\right)}$$ where $f(x)$ is the quartic. We attempt to find a simple expression for $f\left(-x_{1}\right)$. We know that $$f\left(-x_{1}\right)-f\left(x_{1}\right)=2 a \cdot x_{1}^{3}+2 a \cdot x_{1}$$ Since $x_{1}$ is a root, we have $$f\left(-x_{1}\right)=2 a \cdot x_{1}^{3}+2 a \cdot x_{1}$$ Plugging this into our previous expression: $$\frac{x_{4}^{2}}{x_{1}^{2}} \cdot \frac{x_{1}^{3}+x_{1}}{x_{4}^{3}+x_{4}} \frac{x_{1}+\frac{1}{x_{1}}}{x_{4}+\frac{1}{x_{4}}}$$ The expression $x+\frac{1}{x}$ is maximized at $x=2, \frac{1}{2}$ and minimized at $x=1$. We can therefore maximize the numerator with $x_{1}=2$ and minimize the denominator with $x_{4}=1$ to achieve the answer of $\frac{5}{4}$. It can be confirmed that such an answer can be achieved such as with $x_{2}=x_{3}=\frac{\sqrt{10}-1}{3}$.
|
\frac{5}{4}
|
HMMT_2
|
[
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5
|
Determine the largest integer $n$ such that $7^{2048}-1$ is divisible by $2^{n}$.
|
We have $$7^{2048}-1=(7-1)(7+1)\left(7^{2}+1\right)\left(7^{4}+1\right) \cdots\left(7^{1024}+1\right)$$ In the expansion, the eleven terms other than $7+1$ are divisible by 2 exactly once, as can be checked easily with modulo 4.
|
14
|
HMMT_2
|
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5
|
For how many integer values of $b$ does there exist a polynomial function with integer coefficients such that $f(2)=2010$ and $f(b)=8$?
|
We can take $f(x)=-\frac{2002}{d}(x-b)+2010$ for all divisors $d$ of -2002. To see that we can't get any others, note that $b-2$ must divide $f(b)-f(2)$, so $b-2$ divides -2002 (this is because $b-2$ divides $b^{n}-2^{n}$ and hence any sum of numbers of the form $b^{n}-2^{n}$).
|
32
|
HMMT_11
|
[
"Mathematics -> Applied Mathematics -> Probability -> Other",
"Mathematics -> Algebra -> Other"
] | 5
|
In general, if there are $d$ doors in every room (but still only 1 correct door) and $r$ rooms, the last of which leads into Bowser's level, what is the expected number of doors through which Mario will pass before he reaches Bowser's level?
|
Let $E_{i}$ be the expected number of doors through which Mario will pass in the future if he is currently in room $i$ for $i=1,2, \ldots, r+1$ (we will set $E_{r+1}=0$). We claim that $E_{i}=1+\frac{d-1}{d} E_{1}+\frac{1}{d} E_{i+1}$. This is because, as before, there is a $\frac{d-1}{d}$ chance of ending up in room 1, and a $\frac{1}{d}$ chance of ending up in room $i+1$. Note that we can re-write this equation as $d E_{i}=d+(d-1) E_{1}+E_{i+1}$ We can solve this system of equation as follows: let $E_{1}=E_{i}+c_{i}$. Then we can re-write each equation as $d E_{1}-d \cdot c_{i}=d+(d-1) E_{1}+E_{1}-c_{i+1}$, so $c_{i+1}=d \cdot c_{i}+d$, and $c_{1}=0$. We can then see that $c_{i}=d \frac{d^{i-1}-1}{d-1}$ (either by finding the pattern or by using more advanced techniques, such as those given at http://en.wikipedia.org/wiki/Recurrence_relation). Solving for $E_{1}$ using $E_{1}=E_{r}+c_{r}$ and $E_{r}=1+\frac{d-1}{d} E_{1}$, we get $E_{1}=\frac{d\left(d^{r}-1\right)}{d-1}$.
|
\frac{d\left(d^{r}-1\right)}{d-1}
|
HMMT_11
|
[
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Permutations"
] | 5
|
Find all odd positive integers $n>1$ such that there is a permutation $a_{1}, a_{2}, \ldots, a_{n}$ of the numbers $1,2, \ldots, n$, where $n$ divides one of the numbers $a_{k}^{2}-a_{k+1}-1$ and $a_{k}^{2}-a_{k+1}+1$ for each $k, 1 \leq k \leq n$ (we assume $a_{n+1}=a_{1}$ ).
|
Since $\{a_{1}, a_{2}, \ldots, a_{n}\}=\{1,2, \ldots, n\}$ we conclude that $a_{i}-a_{j}$ : $n$ only if $i=j$. From the problem conditions it follows that $$a_{k+1}=a_{k}^{2}+\varepsilon_{k}-n b_{k}$$ where $b_{k} \in \mathbb{Z}$ and $\varepsilon_{k}= \pm 1$. We have $a_{k+1}-a_{l+1}=\left(a_{k}-a_{l}\right)\left(a_{k}+a_{l}\right)+\left(\varepsilon_{k}-\varepsilon_{l}\right)-n\left(b_{k}-b_{l}\right)$. It follows that if $a_{k}+a_{l}=n$ then $\varepsilon_{k} \neq \varepsilon_{l}$ otherwise $a_{k+1}-a_{l+1} \vdots n$ - contradiction. The condition $\varepsilon_{k} \neq \varepsilon_{l}$ means that $\varepsilon_{k}=-\varepsilon_{l}$. Further, one of the $a_{i}$ equals $n$. Let, say, $a_{m}=n$. Then the set $\{a_{1}, a_{2}, \ldots, a_{n}\} \backslash\{a_{m}\}$ can be divided into $\frac{n-1}{2}$ pairs $\left(a_{k}, a_{l}\right)$ such that $a_{k}+a_{l}=n$. For any such pairs of indices $k, l$ we have $\varepsilon_{k}+\varepsilon_{l}=0$. Now add all the equalities for $k=1,2, \ldots, n$. Then $\sum_{k=2}^{n+1} a_{k}=\sum_{k=1}^{n} a_{k}^{2}-n \sum_{k=1}^{n} b_{k}+\varepsilon_{m}$, or $1+2+\ldots+n=1^{2}+2^{2}+\ldots+n^{2}-n \sum_{k=1}^{n} b_{k}+\varepsilon_{m}$ whence $$n \sum_{k=1}^{n} b_{k}=\frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}+\varepsilon_{m}=\frac{n(n+1)(n-1)}{3}+\varepsilon_{m}$$ Note that if $n$ is not divisible by 3 then the number $\frac{n(n+1)(n-1)}{3}$ is divisible by $n$ (since $\frac{(n+1)(n-1)}{3}$ is integer). It follows that $\varepsilon_{m} \vdots n$ which is impossible. Hence $n$ is divisible by 3 and it follows that $\varepsilon_{m}$ is divisible by the number $\frac{n}{3}$. The latter is possible only for $n=3$ because $\varepsilon_{m}= \pm 1$. It remains to verify that $n=3$ satisfies the problem conditions. Indeed, let $a_{1}=1, a_{2}=2, a_{3}=3$. Then $a_{1}^{2}-a_{2}+1=0 \vdots 3, a_{2}^{2}-a_{3}-1=0 \vdots 3$ and $a_{3}^{2}-a_{1}+1=9 \vdots 3$.
|
n=3
|
izho
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
8 students are practicing for a math contest, and they divide into pairs to take a practice test. In how many ways can they be split up?
|
We create the pairs one at a time. The first person has 7 possible partners. Set this pair aside. Of the remaining six people, pick a person. He or she has 5 possible partners. Set this pair aside. Of the remaining four people, pick a person. He or she has 3 possible partners. Set this pair aside. Then the last two must be partners. So there are $7 \cdot 5 \cdot 3=105$ possible groupings. Alternatively, we can consider the 8! permutations of the students in a line, where the first two are a pair, the next two are a pair, etc. Given a grouping, there are 4! ways to arrange the four pairs in order, and in each pair, 2 ways to order the students. So our answer is $\frac{8!}{4!2^{4}}=7 \cdot 5 \cdot 3=105$.
|
105
|
HMMT_11
|
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5
|
Let $ABC$ be a right triangle with hypotenuse $AC$. Let $B^{\prime}$ be the reflection of point $B$ across $AC$, and let $C^{\prime}$ be the reflection of $C$ across $AB^{\prime}$. Find the ratio of $[BCB^{\prime}]$ to $[BC^{\prime}B^{\prime}]$.
|
Since $C, B^{\prime}$, and $C^{\prime}$ are collinear, it is evident that $[BCB^{\prime}]=\frac{1}{2}[BCC^{\prime}]$. It immediately follows that $[BCB^{\prime}]=[BC^{\prime}B^{\prime}]$. Thus, the ratio is 1.
|
1
|
HMMT_2
|
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5
|
Each cell of a $3 \times 3$ grid is labeled with a digit in the set $\{1,2,3,4,5\}$. Then, the maximum entry in each row and each column is recorded. Compute the number of labelings for which every digit from 1 to 5 is recorded at least once.
|
We perform casework by placing the entries from largest to smallest. - The grid must have exactly one 5 since an entry equal to 5 will be the maximum in its row and in its column. We can place this in 9 ways. - An entry equal to 4 must be in the same row or column as the 5; otherwise, it will be recorded twice, so we only have two records left but 1,2, and 3 are all unrecorded. Using similar logic, there is at most one 4 in the grid. So there are 4 ways to place the 4. - We further split into cases for the 3 entries. Without loss of generality, say the 4 and the 5 are in the same row. - If there is a 3 in the same row as the 4 and the 5, then it remains to label a $2 \times 3$ grid with 1s and 2s such that there is exactly one row with all 1s, of which there are $2\left(2^{3}-1\right)=14$ ways to do so. - Suppose there is no 3 in the same row as the 4 and the 5. Then there are two remaining empty rows to place a 3. There are two possible places we could have a record of 2, the remaining unoccupied row or the remaining unoccupied column. There are 2 ways to pick one of these; without loss of generality, we pick the row. Then the column must be filled with all 1s, and the remaining slots in the row with record 2 can be filled in one of 3 ways $(12,21$, or 22$)$. The final empty cell can be filled with a 1,2, or 3, for a total of 3 ways. Our total here is $2 \cdot 2 \cdot 3 \cdot 5=60$ ways. Hence, our final answer is $9 \cdot 4 \cdot(14+60)=36 \cdot 74=2664$.
|
2664
|
HMMT_2
|
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