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 $a$ and $c$ are both between 1 and 9 (from condition 1), there are $\boxed{009}$ possible values (since all digits except $9$ can be expressed this way).

2002 AIME II (Problems • Answer Key • Resources)
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2002 AIME II Problems/Problem 1

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