Let $\phi(n,m), m \neq 1$, be the number of positive integers less than or equal to $n$ that are coprime with $m.$ Clearly, $\phi(m,m) = \phi(m)$, where $\phi(m)$ is Euler’s phi function. Find all integers $m$ that satisfy the following inequality: \[\frac{\phi(n,m)}{n} \geq \frac{\phi(m)}{m}\] for every positive integer $n.$

Show that, for any fixed integer $\,n \geq 1,\,$ the sequence \[ 2, \; 2^2, \; 2^{2^2}, \; 2^{2^{2^2}}, \ldots (\mbox{mod} \; n) \] is eventually constant. [The tower of exponents is defined by $a_1 = 2, \; a_{i+1} = 2^{a_i}$. Also $a_i \; (\mbox{mod} \; n)$ means the remainder which results from dividing $a_i$ by $n$.]

Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2$. The minimum value of the area of $U_1$ can be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Given $a_0 > 1$, the sequence $a_0, a_1, a_2, ...$ is such that for all $k > 0$, $a_k$ is the smallest integer greater than $a_{k-1}$ which is relatively prime to all the earlier terms in the sequence. Find all $a_0$ for which all terms of the sequence are primes or prime powers.

Let $ n, m$ be positive integers of different parity, and $ n > m$. Find all integers $ x$ such that $ \frac {x^{2^n} \minus{} 1}{x^{2^m} \minus{} 1}$ is a perfect square.

For positive integers $a$, $b$, and $c$, define \[ f(a,b,c)=\frac{abc}{\text{gcd}(a,b,c)\cdot\text{lcm}(a,b,c)}. \] We say that a positive integer $n$ is $f@$ if there exist pairwise distinct positive integers $x,y,z\leq60$ that satisfy $f(x,y,z)=n$. How many $f@$ integers are there? [i]Proposed by Michael Ren[/i]

Let $x_{1}$ and $x_{2}$ be relatively prime positive integers. For $n \ge 2$, define $x_{n+1}=x_{n}x_{n-1}+1$.[list=a][*] Prove that for every $i>1$, there exists $j>i$ such that ${x_{i}}^{i}$ divides ${x_{j}}^{j}$. [*] Is it true that $x_{1}$ must divide ${x_{j}}^{j}$ for some $j>1$? [/list]

For positive rational $x$, if $x$ is written in the form $p/q$ with $p, q$ positive relatively prime integers, define $f(x)=p+q$. For example, $f(1)=2$. a) Prove that if $f(x)=f(mx/n)$ for rational $x$ and positive integers $m, n$, then $f(x)$ divides $|m-n|$. b) Let $n$ be a positive integer. If all $x$ which satisfy $f(x)=f(2^nx)$ also satisfy $f(x)=2^n-1$, find all possible values of $n$. [i]Anderson Wang.[/i]

Let $\vartriangle ABC$ be a triangle. Let $Q$ be a point in the interior of $\vartriangle ABC$, and let $X, Y,Z$ denote the feet of the altitudes from $Q$ to sides $BC$, $CA$, $AB$, respectively. Suppose that $BC = 15$, $\angle ABC = 60^o$, $BZ = 8$, $ZQ = 6$, and $\angle QCA = 30^o$. Let line $QX$ intersect the circumcircle of $\vartriangle XY Z$ at the point $W\ne X$. If the ratio $\frac{ WY}{WZ}$ can be expressed as $\frac{p}{q}$ for relatively prime positive integers $p, q$, find $p + q$.

One of the boxes that Joshua and Wendy unpack has Joshua's collection of board games. Michael, Wendy, Alexis, and Joshua decide to play one of them, a game called $\textit{Risk}$ that involves rolling ordinary six-sided dice to determine the outcomes of strategic battles. Wendy has never played before, so early on Michael explains a bit of strategy. "You have the first move and you occupy three of the four territories in the Australian continent. You'll want to attack Joshua in Indonesia so that you can claim the Australian continent which will give you bonus armies on your next turn." "Don't tell her $\textit{that!}$" complains Joshua. Wendy and Joshua begin rolling dice to determine the outcome of their struggle over Indonesia. Joshua rolls extremely well, overcoming longshot odds to hold off Wendy's attack. Finally, Wendy is left with one chance. Wendy and Joshua each roll just one six-sided die. Wendy wins if her roll is $\textit{higher}$ than Joshua's roll. Let $a$ and $b$ be relatively prime positive integers so that $a/b$ is the probability that Wendy rolls higher, giving her control over the continent of Australia. Find the value of $a+b$.

As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F. $ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form \[\frac{a}{b}\cdot\pi-\sqrt{c}+d,\] where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$? [asy] size(6cm); filldraw(circle((0,0),2), gray(0.7)); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label("$A$",(-3,-1),S); dot((-2,0)); label("$E$",(-2,0),NW); dot((-1,-1)); label("$B$",(-1,-1),S); dot((0,0)); label("$F$",(0,0),N); dot((1,-1)); label("$C$",(1,-1), S); dot((2,0)); label("$G$", (2,0),NE); dot((3,-1)); label("$D$", (3,-1), S); [/asy] $\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$

A quadratic trinomial $p(x)$ with real coefficients is given. Prove that there is a positive integer $n$ such that the equation $p(x) = \frac{1}{n}$ has no rational roots.

A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum-shaped solid $F$, in such a way that the ratio between the areas of the painted surfaces of $C$ and $F$ and the ratio between the volumes of $C$ and $F$ are both equal to $k$. Given that $k=m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. The ratio of the area of square $EFGH$ to the area of square $ABCD$ can be expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m<n$. Find $10n+m$.

In triangle $ABC$, $AB=13$, $BC=15$, and $CA=14$. Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$

Prove that if $x,y$ and $z$ are pairwise relatively prime positive integers, and if $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$, then $x+y, x-z, y-z$ are perfect squares of integers.

The diagram below shows twelve $30-60-90$ triangles placed in a circle so that the hypotenuse of each triangle coincides with the longer leg of the next triangle. The fourth and last triangle in this diagram are shaded. The ratio of the perimeters of these two triangles can be written as $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] size(200); defaultpen(linewidth(0.8)); pair point=(-sqrt(3),0); pair past,unit; path line; for(int i=0;i<=12;++i) { past = point; line=past--origin; unit=waypoint(line,1/200); point=extension(past,rotate(90,past)*unit,origin,dir(180-30*i)); if (i == 4) { filldraw(origin--past--point--cycle,gray(0.7)); } else if (i==12) { filldraw(origin--past--point--cycle,gray(0.7)); } else { draw(origin--past--point); } } draw(origin--point); [/asy]

For each natural number $a$ we denote $\tau (a)$ and $\phi (a)$ the number of natural numbers dividing $a$ and the number of natural numbers less than $a$ that are relatively prime to $a$. Find all natural numbers $n$ for which $n$ has exactly two different prime divisors and $n$ satisfies $\tau (\phi (n))=\phi (\tau (n))$.

Find a pair of relatively prime four digit natural numbers $A$ and $B$ such that for all natural numbers $m$ and $n$, $\vert A^m -B^n \vert \ge 400$.

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD$. The crease is $EF$, where $E$ is on $AB$ and $F$is on $CD$. The dimensions $AE=8$, $BE=17$, and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

$\tfrac11+\tfrac13+\tfrac15=\tfrac12+\tfrac14+\tfrac16+\tfrac m n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Let $ S \equal{} \{1,2,3,\cdots ,280\}$. Find the smallest integer $ n$ such that each $ n$-element subset of $ S$ contains five numbers which are pairwise relatively prime.

A person flips $2010$ coins at a time. He gains one penny every time he flips a prime number of heads, but must stop once he flips a non-prime number. If his expected amount of money gained in dollars is $\frac{a}{b}$, where $a$ and $b$ are relatively prime, compute $\lceil\log_{2}(100a+b)\rceil$. [i]Proposed by Lewis Chen[/i]

How many reorderings of $2,3,4,5,6$ have the property that every pair of adjacent numbers are relatively prime? [i]2021 CCA Math Bonanza Individual Round #3[/i]

In the expansion of $(ax+b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a+b.$