2002 AIME II Problems/Problem 1

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Problem

Given that

$\begin{eqnarray*}&(1)& \text{x and y are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \q...$

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 $\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\\end{eqnarray*}$

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

2002 AIME II (Problems • Resources)
Preceded by First Question	Followed by Problem 2
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2002 AIME II Problems/Problem 1

From AoPSWiki

Problem

Solution

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