# Difference between revisions of "2002 AIME II Problems/Problem 1"

## Problem

Given that $\begin{eqnarray*}&(1)& x\text{ and }y\text{ are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\ &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\ &(3)& z=|x-y|. \end{eqnarray*}$ How many distinct values of $z$ are possible?

## Solution

We express the numbers as $x=100a+10b+c$ and $y=100c+10b+a$. From this, we have $\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\ \end{eqnarray*}$ Because $a$ and $c$ are digits, and $a$ is between 1 and 9, there are $\boxed{9}$ possible values.

## See also

 2002 AIME II (Problems • Answer Key • Resources) Preceded byFirst Question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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