2002 AIME II Problems/Problem 1

Revision as of 15:14, 9 August 2008 by Mathgeek2006 (talk | contribs) (Solution)

Problem

Given that

$\begin{eqnarray*}&(1)& \text{x and y are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\
   &(2)& \text{y is the number formed by reversing the digits of x; and}\\
&(3)& z=|x-y|. \end{eqnarray*}$ (Error compiling LaTeX. ! Missing \endgroup inserted.)

How many distinct values of $z$ are possible?

Solution

We express the numbers as $x=100a+10b+c$ and $100c+10b+a$. From this, we have

$\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\ \end{eqnarray*}$ (Error compiling LaTeX. ! Missing \endgroup inserted.)

Because $a$ and $c$ are digits, and $a$ is between 1 and 9, there are 9 possible values.

See also

2002 AIME II (ProblemsAnswer KeyResources)
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
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