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, $c$ is greater than 0, and $a$ is between 1 and 9, there are $\boxed{8}$ possible values.

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

Page

Toolbox

Search

2002 AIME II Problems/Problem 1

Problem

Solution

See also