# Difference between revisions of "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 number is divisible by $5$, This gives us a answer of $10\times10\times10\times\frac{1}{5}=\boxed{200}$ - mathleticguyyy