This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 8

2010 Putnam, A4

Prove that for each positive integer $n,$ the number $10^{10^{10^n}}+10^{10^n}+10^n-1$ is not prime.

1998 Putnam, 4

Let $A_1=0$ and $A_2=1$. For $n>2$, the number $A_n$ is defined by concatenating the decimal expansions of $A_{n-1}$ and $A_{n-2}$ from left to right. For example $A_3=A_2A_1=10$, $A_4=A_3A_2=101$, $A_5=A_4A_3=10110$, and so forth. Determine all $n$ such that $11$ divides $A_n$.

1999 Putnam, 6

Let $S$ be a finite set of integers, each greater than $1$. Suppose that for each integer $n$ there is some $s\in S$ such that $\gcd(s,n)=1$ or $\gcd(s,n)=s$. Show that there exist $s,t\in S$ such that $\gcd(s,t)$ is prime.

2015 Putnam, A3

Compute \[\log_2\left(\prod_{a=1}^{2015}\prod_{b=1}^{2015}\left(1+e^{2\pi iab/2015}\right)\right)\] Here $i$ is the imaginary unit (that is, $i^2=-1$).

2015 Putnam, A5

Let $q$ be an odd positive integer, and let $N_q$ denote the number of integers $a$ such that $0<a<q/4$ and $\gcd(a,q)=1.$ Show that $N_q$ is odd if and only if $q$ is of the form $p^k$ with $k$ a positive integer and $p$ a prime congruent to $5$ or $7$ modulo $8.$

2015 Putnam, B6

For each positive integer $k,$ let $A(k)$ be the number of odd divisors of $k$ in the interval $\left[1,\sqrt{2k}\right).$ Evaluate: \[\sum_{k=1}^{\infty}(-1)^{k-1}\frac{A(k)}k.\]

Russian TST 2016, P2

Let $q$ be an odd positive integer, and let $N_q$ denote the number of integers $a$ such that $0<a<q/4$ and $\gcd(a,q)=1.$ Show that $N_q$ is odd if and only if $q$ is of the form $p^k$ with $k$ a positive integer and $p$ a prime congruent to $5$ or $7$ modulo $8.$

2016 Putnam, A1

Find the smallest positive integer $j$ such that for every polynomial $p(x)$ with integer coefficients and for every integer $k,$ the integer \[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\] (the $j$-th derivative of $p(x)$ at $k$) is divisible by $2016.$