Let $T=\text{TNFTPP}$. Fermi and Feynman play the game $\textit{Probabicloneme}$ in which Fermi wins with probability $a/b$, where $a$ and $b$ are relatively prime positive integers such that $a/b<1/2$. The rest of the time Feynman wins (there are no ties or incomplete games). It takes a negligible amount of time for the two geniuses to play $\textit{Probabicloneme}$ so they play many many times. Assuming they can play infinitely many games (eh, they're in Physicist Heaven, we can bend the rules), the probability that they are ever tied in total wins after they start (they have the same positive win totals) is $(T-332)/(2T-601)$. Find the value of $a$.

Let $f(x)=x^2+4x+2$. Let $r$ be the difference between the largest and smallest real solutions of the equation $f(f(f(f(x))))=0$. Then $r=a^{\frac{p}{q}}$ for some positive integers $a$, $p$, $q$ so $a$ is square-free and $p,q$ are relatively prime positive integers. Compute $a+p+q$.

Let $ABCD$ be a square with side length $2$. Let $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{CD}$ respectively, and let $X$ and $Y$ be the feet of the perpendiculars from $A$ to $\overline{MD}$ and $\overline{NB}$, also respectively. The square of the length of segment $\overline{XY}$ can be written in the form $\tfrac pq$ where $p$ and $q$ are positive relatively prime integers. What is $100p+q$? [i]Proposed by David Altizio[/i]

The diagram shows a large circular dart board with four smaller shaded circles each internally tangent to the larger circle. Two of the internal circles have half the radius of the large circle, and are, therefore, tangent to each other. The other two smaller circles are tangent to these circles. If a dart is thrown so that it sticks to a point randomly chosen on the dart board, then the probability that the dart sticks to a point in the shaded area is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] size(150); defaultpen(linewidth(0.8)); filldraw(circle((0,0.5),.5),gray); filldraw(circle((0,-0.5),.5),gray); filldraw(circle((2/3,0),1/3),gray); filldraw(circle((-2/3,0),1/3),gray); draw(unitcircle); [/asy]

Square $ABCD$ is shown in the diagram below. Points $E$, $F$, and $G$ are on sides $\overline{AB}$, $\overline{BC}$ and $\overline{DA}$, respectively, so that lengths $\overline{BE}$, $\overline{BF}$, and $\overline{DG}$ are equal. Points $H$ and $I$ are the midpoints of segments $\overline{EF}$ and $\overline{CG}$, respectively. Segment $\overline{GJ}$ is the perpendicular bisector of segment $\overline{HI}$. The ratio of the areas of pentagon $AEHJG$ and quadrilateral $CIHF$ can be written as $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] draw((0,0)--(50,0)--(50,50)--(0,50)--cycle); label("$A$",(0,50),NW); label("$B$",(50,50),NE); label("$C$",(50,0),SE); label("$D$",(0,0),SW); label("$E$",(0,100/3-1),W); label("$F$",(100/3-1,0),S); label("$G$",(20,50),N); label("$H$",((100/3-1)/2,(100/3-1)/2),SW); label("$I$",(35,25),NE); label("$J$",(((100/3-1)/2+35)/2,((100/3-1)/2+25)/2),S); draw((0,100/3-1)--(100/3-1,0)); draw((20,50)--(50,0)); draw((100/6-1/2,100/6-1/2)--(35,25)); draw((((100/3-1)/2+35)/2,((100/3-1)/2+25)/2)--(20,50)); [/asy]

For a positive integer $n$, define the set $S_n$ as $S_n =\{(a, b)|a, b \in N, lcm[a, b] = n\}$ . Let $f(n)$ be the sum of $\phi (a)\phi (b)$ for all $(a, b) \in S_n$. If a prime $p$ relatively prime to $n$ is a divisor of $f(n)$, prove that there exists a prime $q|n$ such that $p|q^2 - 1$.

[b]Problem:[/b]For a positive integer $ n$,let $ V(n; b)$ be the number of decompositions of $ n$ into a product of one or more positive integers greater than $ b$. For example,$ 36 \equal{} 6.6 \equal{}4.9 \equal{} 3.12 \equal{} 3 .3. 4$, so that $ V(36; 2) \equal{} 5$.Prove that for all positive integers $ n$; b it holds that $ V(n;b)<\frac{n}{b}$. :)

Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Find all positive integers $n$ with the property that the set \[\{n,n+1,n+2,n+3,n+4,n+5\}\] can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.

For all positive integers $n$, prove that there are integers $x, y$ relatively prime to $5$ such that $x^2 + y^2 = 5^n$.

Let $m$ boxes be given, with some balls in each box. Let $n < m$ be a given integer. The following operation is performed: choose $n$ of the boxes and put $1$ ball in each of them. Prove: [i](a) [/i]If $m$ and $n$ are relatively prime, then it is possible, by performing the operation a finite number of times, to arrive at the situation that all the boxes contain an equal number of balls. [i](b)[/i] If $m$ and $n$ are not relatively prime, there exist initial distributions of balls in the boxes such that an equal distribution is not possible to achieve.

If an integer $a > 1$ is given such that $\left(a-1\right)^3+a^3+\left(a+1\right)^3$ is the cube of an integer, then show that $4\mid a$.

For every odd number $p>1$ we have: $\textbf{(A)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-2\qquad \textbf{(B)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p\\ \textbf{(C)}\ (p-1)^{\frac{1}{2}(p-1)} \; \text{is divisible by} \; p\qquad \textbf{(D)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p+1\\ \textbf{(E)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-1$

Suppose $x$ is a random real number between $1$ and $4$, and $y$ is a random real number between $1$ and $9$. If the expected value of \[ \left\lceil \log_2 x \right\rceil - \left\lfloor \log_3 y \right\rfloor \] can be expressed as $\frac mn$ where $m$ and $n$ are relatively prime positive integers, compute $100m + n$. [i]Proposed by Lewis Chen[/i]

Let $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from 1 to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer.

Each of $27$ bricks (right rectangular prisms) has dimensions $a \times b \times c$, where $a$, $b$, and $c$ are pairwise relatively prime positive integers. These bricks are arranged to form a $3 \times 3 \times 3$ block, as shown on the left below. A $28$[sup]th[/sup] brick with the same dimensions is introduced, and these bricks are reconfigured into a $2 \times 2 \times 7$ block, shown on the right. The new block is $1$ unit taller, $1$ unit wider, and $1$ unit deeper than the old one. What is $a + b + c$? [img]https://cdn.artofproblemsolving.com/attachments/2/d/b18d3d0a9e5005c889b34e79c6dab3aaefeffd.png[/img] $ \textbf{(A) }88 \qquad \textbf{(B) }89 \qquad \textbf{(C) }90 \qquad \textbf{(D) }91 \qquad \textbf{(E) }92 \qquad $

How many primes less than $100$ have $7$ as the ones digit? (Assume the usual base ten representation) $\text{(A)} \ 4 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 7 \qquad \text{(E)} \ 8$

Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Triangle $AB_0C_0$ has side lengths $AB_0 = 12$, $B_0C_0 = 17$, and $C_0A = 25$. For each positive integer $n$, points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$, respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$. The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$, where $p$ and $q$ are relatively prime positive integers. Find $q$.

Find $100m+n$ if $m$ and $n$ are relatively prime positive integers such that \[ \sum_{\substack{i,j \ge 0 \\ i+j \text{ odd}}} \frac{1}{2^i3^j} = \frac{m}{n}. \][i]Proposed by Aaron Lin[/i]

$(FRA 6)$ Consider the integer $d = \frac{a^b-1}{c}$, where $a, b$, and $c$ are positive integers and $c \le a.$ Prove that the set $G$ of integers that are between $1$ and $d$ and relatively prime to $d$ (the number of such integers is denoted by $\phi(d)$) can be partitioned into $n$ subsets, each of which consists of $b$ elements. What can be said about the rational number $\frac{\phi(d)}{b}?$

When rolling a certain unfair six-sided die with faces numbered $1, 2, 3, 4, 5$, and $6$, the probability of obtaining face $F$ is greater than $\frac{1}{6}$, the probability of obtaining the face opposite is less than $\frac{1}{6}$, the probability of obtaining any one of the other four faces is $\frac{1}{6}$, and the sum of the numbers on opposite faces is $7$. When two such dice are rolled, the probability of obtaining a sum of $7$ is $\frac{47}{288}$. Given that the probability of obtaining face $F$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Given a fixed positive integer $a\geq 9$. Prove: There exist finitely many positive integers $n$, satisfying: (1)$\tau (n)=a$ (2)$n|\phi (n)+\sigma (n)$ Note: For positive integer $n$, $\tau (n)$ is the number of positive divisors of $n$, $\phi (n)$ is the number of positive integers $\leq n$ and relatively prime with $n$, $\sigma (n)$ is the sum of positive divisors of $n$.

Let $a$ be a rational number and let $n$ be a positive integer. Prove that the polynomial $X^{2^n}(X+a)^{2^n}+1$ is irreducible in the ring $\mathbb{Q}[X]$ of polynomials with rational coefficients. [i]Proposed by Vincent Jugé, École Polytechnique, Paris.[/i]

The sum $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$