Find all positive integers $n$ such that the equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ has exactly $2011$ positive integer solutions $(x,y)$ where $x \leq y$.

Let $n$ be a given number greater than 2. We consider the set $V_n$ of all the integers of the form $1 + kn$ with $k = 1, 2, \ldots$ A number $m$ from $V_n$ is called indecomposable in $V_n$ if there are not two numbers $p$ and $q$ from $V_n$ so that $m = pq.$ Prove that there exist a number $r \in V_n$ that can be expressed as the product of elements indecomposable in $V_n$ in more than one way. (Expressions which differ only in order of the elements of $V_n$ will be considered the same.)

To every natural number $k, k \geq 2$, there corresponds a sequence $a_n(k)$ according to the following rule: \[a_0 = k, \qquad a_n = \tau(a_{n-1}) \quad \forall n \geq 1,\] in which $\tau(a)$ is the number of different divisors of $a$. Find all $k$ for which the sequence $a_n(k)$ does not contain the square of an integer.

Prove that there exist infinitely many positive integers which cannot be written in form $a^{d(a)}+b^{d(b)}$ for some positive integers $a$ and $b$ For positive integer $d(a)$ denotes number of positive divisors of $a$ [i]Proposed by Borna Vukorepa[/i]

How many zeros are at the end of the product \[25\times 25\times 25\times 25\times 25\times 25\times 25\times 8\times 8\times 8?\] $\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

For a finite set $X$ define \[ S(X) = \sum_{x \in X} x \text{ and } P(x) = \prod_{x \in X} x. \] Let $A$ and $B$ be two finite sets of positive integers such that $\left\lvert A \right\rvert = \left\lvert B \right\rvert$, $P(A) = P(B)$ and $S(A) \neq S(B)$. Suppose for any $n \in A \cup B$ and prime $p$ dividing $n$, we have $p^{36} \mid n$ and $p^{37} \nmid n$. Prove that \[ \left\lvert S(A) - S(B) \right\rvert > 1.9 \cdot 10^{6}. \][i]Proposed by Evan Chen[/i]

Is it possible to place $1995$ different natural numbers along a circle so that for any two of these numbers, the ratio of the greatest to the least is a prime? I feel that my solution's wording and notation is awkward (and perhaps unnecessarily complicated), so please feel free to critique it: [hide] Suppose that we do have such a configuration $a_{1},a_{2},...a_{1995}$. WLOG, $a_{2}=p_{1}a_{1}$. Then \[\frac{a_{2}}{a_{3}}= p_{2}, \frac{1}{p_{2}}\] \[\frac{a_{3}}{a_{4}}= p_{3}, \frac{1}{p_{3}}\] \[... \] \[\frac{a_{1995}}{a_{1}}= p_{1995}, \frac{1}{p_{1995}}\] Multiplying these all together, \[\frac{a_{2}}{a_{1}}= \frac{\prod p_{k}}{\prod p_{j}}= p_{1}\] Where $\prod p_{k}$ is some product of the elements in a subset of $\{ p_{2},p_{3}, ...p_{1995}\}$. We clear denominators to get \[p_{1}\prod p_{j}= \prod p_{k}\] Now, by unique prime factorization, the set $\{ p_{j}\}\cup \{ p_{1}\}$ is equal to the set $\{ p_{k}\}$. However, since there are a total of $1995$ primes, this is impossible. We conclude that no such configuration exists. [/hide]

Let $A$ be the greatest possible value of a product of positive integers that sums to $2014$. Compute the sum of all bases and exponents in the prime factorization of $A$. For example, if $A=7\cdot 11^5$, the answer would be $7+11+5=23$.

Let $\eta(m)$ be the product of all positive integers that divide $m$, including $1$ and $m$. If $\eta(\eta(\eta(10))) = 10^n$, compute $n$. [i]Proposed by Kevin Sun[/i]

For any positive integer $n$, let $\tau (n)$ denote the number of its positive divisors (including 1 and itself). Determine all positive integers $m$ for which there exists a positive integer $n$ such that $\frac{\tau (n^{2})}{\tau (n)}=m$.

If $ a$, $ b$, and $ c$ are positive real numbers such that $ a(b \plus{} c) \equal{} 152$, $ b(c \plus{} a) \equal{} 162$, and $ c(a \plus{} b) \equal{} 170$, then abc is $ \textbf{(A)}\ 672 \qquad \textbf{(B)}\ 688 \qquad \textbf{(C)}\ 704 \qquad \textbf{(D)}\ 720 \qquad \textbf{(E)}\ 750$

Find all the integer pairs $ (x,y)$ such that: \[ x^2\minus{}y^4\equal{}2009\]

What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1,k_2,\ldots,k_n$ for which \[k_1^2+k_2^2+\ldots+k_n^2=2002?\] $\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18$

Consider a sequence whose first term is a given positive integer \( N > 1 \). Consider the prime factorization of \( N \). If \( N \) is a power of 2, the sequence consists of a single term: \( N \). Otherwise, the second term of the sequence is obtained by replacing the largest prime factor \( p \) of \( N \) with \( p + 1 \) in the prime factorization. If the new number is not a power of 2, we repeat the same procedure with it, remembering to factor it again into primes. If it is a power of 2, the numerical sequence ends. And so on. For example, if the first term of the sequence is \( N = 300 = 2^2 \cdot 3 \cdot 5^2 \), since its largest prime factor is \( p = 5 \), the second term is \( 2^2 \cdot 3 \cdot (5 + 1)^2 = 2^4 \cdot 3^3 \). Repeating the procedure, the largest prime factor of the second term is \( p = 3 \), so the third term is \( 2^4 \cdot (3 + 1)^3 = 2^{10} \). Since we obtained a power of 2, the sequence has 3 terms: \( 2^2 \cdot 3 \cdot 5^2 \), \( 2^4 \cdot 3^3 \), and \( 2^{10} \). a) How many terms does the sequence have if the first term is \( N = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \)? b) Show that if a prime factor \( p \) leaves a remainder of 1 when divided by 3, then \( \frac{p+1}{2} \) is an integer that also leaves a remainder of 1 when divided by 3. c) Present an initial term \( N \) less than 1,000,000 (one million) such that the sequence starting from \( N \) has exactly 11 terms.

An integer $n$ is called [i]apocalyptic[/i] if the addition of $6$ different positive divisors of $n$ gives $3528$. For example, $2012$ is apocalyptic, because it has six divisors, $1$, $2$, $4$, $503$, $1006$ and $2012$, that add up to $3528$. Find the smallest positive apocalyptic number.

Find all pairs of positive integers $ x,y$ satisfying the equation \[ y^x \equal{} x^{50}\]

Find four positive integers each not exceeding $70000$ and each having more than $100$ divisors.

Define a function $P(n)$ from the set of positive integers to itself, where $P(1)=1$ and if an integer $n > 1$ has prime factorization $$n = p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$$ then $$P(n) = a_1^{p_1}a_2^{p_2} \dots a_k^{p_k}.$$ Prove that $P(P(n)) \le n$ for all positive integers $n.$ [i]Proposed by Evan Chang (squareman), USA[/i]

What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1,k_2,\ldots,k_n$ for which \[k_1^2+k_2^2+\ldots+k_n^2=2002?\] $\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18$

If $x$ is the cube of a positive integer and $d$ is the number of positive integers that are divisors of $x$, then $d$ could be $ \textbf{(A)}\ 200\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 202\qquad\textbf{(D)}\ 203\qquad\textbf{(E)}\ 204 $

Let $A$ be the greatest possible value of a product of positive integers that sums to $2014$. Compute the sum of all bases and exponents in the prime factorization of $A$. For example, if $A=7\cdot 11^5$, the answer would be $7+11+5=23$.

A positive integer $N$ is [i]interoceanic[/i] if its prime factorization $$N=p_1^{x_1}p_2^{x_2}\cdots p_k^{x_k}$$ satisfies $$x_1+x_2+\dots +x_k=p_1+p_2+\cdots +p_k.$$ Find all interoceanic numbers less than 2020.

For every composite positive integer $n$, define $r(n)$ to be the sum of the factors in the prime factorization of $n$. For example, $r(50)=12$ because the prime factorization of $50$ is $ 2 \cdot 5^2 $, and $ 2 + 5 + 5 = 12 $. What is the range of the function $r$, $ \{ r(n) : n \ \text{is a composite positive integer} \} $? [b](A)[/b] the set of positive integers [b](B)[/b] the set of composite positive integers [b](C)[/b] the set of even positive integers [b](D)[/b] the set of integers greater than 3 [b](E)[/b] the set of integers greater than 4

For a nonzero integer $i$, the exponent of $2$ in the prime factorization of $i$ is called $ord_2 (i)$. For example, $ord_2(9)=0$ since $9$ is odd, and $ord_2(28)=2$ since $28=2^2\times7$. The numbers $3^n-1$ for $n=1,2,3,\ldots$ are all even so $ord_2(3^n-1)>0$ for $n>0$. a) For which positive integers $n$ is $ord_2(3^n-1) = 1$? b) For which positive integers $n$ is $ord_2(3^n-1) = 2$? c) For which positive integers $n$ is $ord_2(3^n-1) = 3$? Prove your answers.

Find all integers $n$, $n \ge 1$, such that $n \cdot 2^{n+1}+1$ is a perfect square.