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

Revision as of 14:14, 9 August 2008

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. Unknown error_msg)

How many distinct values of $z$ are possible?

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. Unknown error_msg)

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

@@ Line 7: / Line 7: @@
 == Solution ==
-We express the numbers as <math>x=100a+10b+c</math> and <math>100c+10b+c</math>.  From this, we have <center><math>\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\
+We express the numbers as <math>x=100a+10b+c</math> and <math>100c+10b+a</math>.  From this, we have <center><math>\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\
 \end{eqnarray*}</math></center>
 Because <math>a</math> and <math>c</math> are digits, and <math>a</math> is between 1 and 9, there are 9 possible values.

2002 AIME II (Problems • Answer Key • Resources)
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