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: 1340

Let $M$ be an integer, and let $p$ be a prime with $p>25$. Show that the set $\{M, M+1, \cdots, M+ 3\lfloor \sqrt{p} \rfloor -1\}$ contains a quadratic non-residue to modulus $p$.
Let $p>3$ is a prime number and $k=\lfloor\frac{2p}{3}\rfloor$. Prove that \[{p \choose 1}+{p \choose 2}+\cdots+{p \choose k}\] is divisible by $p^{2}$.
Let $x =\sqrt{a}+\sqrt{b}$, where $a$ and $b$ are natural numbers, $x$ is not an integer, and $x < 1976$. Prove that the fractional part of $x$ exceeds $10^{-19.76}$.
The sum of the base-$ 10$ logarithms of the divisors of $ 10^n$ is $ 792$. What is $ n$? $ \textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$
$n$ being a given integer, find all functions $f\colon \mathbb{Z} \to \mathbb{Z}$, such that for all integers $x,y$ we have $f\left( {x + y + f(y)} \right) = f(x) + ny$.
Let $ A$, $ B$ be the number of digits of $ 2^{1998}$ and $ 5^{1998}$ in decimal system. $ A \plus{} B \equal{} ?$ $\textbf{(A)}\ 1998 \qquad\textbf{(B)}\ 1999 \qquad\textbf{(C)}\ 2000 \qquad\textbf{(D)}\ 3996 \qquad\textbf{(E)}\ 3998$
Prove: From the set $\{1,2,...,n\}$, one can choose a subset with at most $2 \left\lfloor \sqrt n \right\rfloor +1$ elements such that the set of the pairwise differences from this subset is $\{1,2,...,n-1\}$. ($\left\lfloor x \right\rfloor$ means the greatest integer $\leq x$)
A convex 2011-gon is drawn on the board. Peter keeps drawing its diagonals in such a way, that each newly drawn diagonal intersected no more than one of the already drawn diagonals. What is the greatest number of diagonals that Peter can draw?
Let $f$ and $g$ be two functions such that \[f(x)=\frac{1}{\lfloor | x | \rfloor}, \quad g(x)=\frac{1}{|\lfloor x \rfloor |}.\] Find the domains of $f$ and $g$ and then prove that \[\lim_{x \to -1^+} f(x)= \lim_{x \to 1^- } g(x).\]
Find the least positive integer $k$ for which the equation $\lfloor \frac{2002}{n}\rfloor = k$ has no integer solutions for $n.$ (The notation $\lfloor x \rfloor$ means the greatest integer less than or equal to $x.$)
Determine whether there exists a polynomial $f(x_1, x_2)$ with two variables, with integer coefficients, and two points $A=(a_1, a_2)$ and $B=(b_1, b_2)$ in the plane, satisfying the following conditions: (i) $A$ is an integer point (i.e $a_1$ and $a_2$ are integers); (ii) $|a_1-b_1|+|a_2-b_2|=2010$; (iii) $f(n_1, n_2)>f(a_1, a_2)$ for all integer points $(n_1, n_2)$ in the plane other than $A$; (iv) $f(x_1, x_2)>f(b_1, b_2)$ for all integer points $(x_1, x_2)$ in the plane other than $B$. [i]Massimo Gobbino, Italy[/i]
Consider the sequence of numbers: $ 4, 7, 1, 8, 9, 7, 6, \ldots .$ For $ n > 2$, the $ n$th term of the sequence is the units digit of the sum of the two previous terms. Let $ S_n$ denote the sum of the first $ n$ terms of this sequence. The smallest value of $ n$ for which $ S_n > 10,000$ is: $ \textbf{(A)}\ 1992 \qquad \textbf{(B)}\ 1999 \qquad \textbf{(C)}\ 2001 \qquad \textbf{(D)}\ 2002 \qquad \textbf{(E)}\ 2004$
Let $a$ and $b$ be two positive integers. Prove that the integer \[a^2+\left\lceil\frac{4a^2}b\right\rceil\] is not a square. (Here $\lceil z\rceil$ denotes the least integer greater than or equal to $z$.) [i]Russia[/i]
Let $ \lfloor x \rfloor$ be the greatest integer less than or equal to $ x$. Then the number of real solutions to $ 4x^2 \minus{} 40 \lfloor x \rfloor \plus{} 51 \equal{} 0$ is $ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
For every positive integer $n$, form the number $n/s(n)$, where $s(n)$ is the sum of digits of $n$ in base 10. Determine the minimum value of $n/s(n)$ in each of the following cases: (i) $10 \leq n \leq 99$ (ii) $100 \leq n \leq 999$ (iii) $1000 \leq n \leq 9999$ (iv) $10000 \leq n \leq 99999$
Solve equation $x^4 = 2x^2 + \lfloor x \rfloor$, where $ \lfloor x \rfloor$ is an integral part of $x$.
For real numbers $a$ and $b$, define the sequence $\{x_{a,b}(n)\}$ as follows: $x_{a,b}(1)=a$, $x_{a,b}(2)=b$, and for $n>1$, $x_{a,b}(n+1)=(x_{a+b}(n-1))^2+(x_{a,b}(n))^2$. For real numbers $c$ and $d$, define the sequence $\{y_{c,d}(n)\}$ as follows: $y_{c,d}(1)=c$, $y_{c,d}(2)=d$, and for $n>1$, $y_{c,d}(n+1)=(y_{c,d}(n-1)+y_{c,d}(n))^2$. Call $(a,b,c)$ a good triple if there exists $d$ such that for all $n$ sufficiently large, $y_{c,d}(n)=(x_{a,b}(n))^2$. For some $(a,b)$ there are exactly three values of $c$ that make $(a,b,c)$ a good triple. Among these pairs $(a,b)$, compute the maximum value of $\lfloor 100(a+b)\rfloor$.
The sequence $(a_n)$ is defined by $a_0 = 0$ and $a_{n+1} = [\sqrt[3]{a_n +n}]^3$ for $n \ge 0$. (a) Find $a_n$ in terms of $n$. (b) Find all $n$ for which $a_n = n$.
Let $S$ be the sum of all $x$ such that $1\leq x\leq 99$ and \[\{x^2\}=\{x\}^2.\] Compute $\lfloor S\rfloor$.
Find all real solutions to the equation $$\lfloor x \rfloor ^2 + \lfloor x \rfloor= x^2-\frac14.$$
If $ n$ runs through all the positive integers, then $ f(n) \equal{} \left \lfloor n \plus{} \sqrt {3n} \plus{} \frac {1}{2} \right \rfloor$ runs through all positive integers skipping the terms of the sequence $ a_n \equal{} \left \lfloor \frac {n^2 \plus{} 2n}{3} \right \rfloor$.
Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $(0,1)$. What is the probability that $\lfloor \log_2{x} \rfloor=\lfloor \log_2{y} \rfloor$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to the real number $r$? $\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$
For every integer $n>2$, prove the equality $$\left\lfloor\frac{n(n+1)}{4n-2}\right\rfloor=\left\lfloor\frac{n+1}4\right\rfloor.$$
Prove that for all $n\in\mathbb{Z}^+$, we have \[ \sum\limits_{p=1}^n\sum\limits_{q=1}^p\left\lfloor -\frac{1+\sqrt{8q+(2p-1)^2}}{2}\right\rfloor =-\frac{n(n+1)(n+2)}{3} \]
Let $D$ be the set of all pairs $(i,j)$, $1\le i,j\le n$. Prove there exists a subset $S \subset D$, with $|S|\ge\left \lfloor\frac{3n(n+1)}{5}\right \rfloor$, such that for any $(x_1,y_1), (x_2,y_2) \in S$ we have $(x_1+x_2,y_1+y_2) \not \in S$. (Peter Cameron)