# 2013 AIME I Problems/Problem 2

## Problem 2

Find the number of five-digit positive integers, $n$, that satisfy the following conditions:

(a) the number $n$ is divisible by $5,$
(b) the first and last digits of $n$ are equal, and
(c) the sum of the digits of $n$ is divisible by $5.$

## Solution

The number takes a form of $5\text{x,y,z}5$, in which $5|x+y+z$. Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$, there are exactly two values of $z$ that satisfy the condition of $5|x+y+z$. Therefore, the answer is $10\times10\times2=\boxed{200}$

## Solution 2

For a number to be divisible by $5$, the last digit of the number must be $5$ or $0$. However, since the first digit is the same as the last one, the last (and first) digits can not be $0$, so the number must be in the form $\overline{5abc5}$, where $a+b+c$ is divisible by 5 Since there is a $\frac{1}{5}$ chance that sum of digits of a randomly selected positive integer is divisible by $5$, This gives us a answer of $10\times10\times10\times\frac{1}{5}=\boxed{200}$ - mathleticguyyy

## See also

 2013 AIME I (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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