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

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== Problem ==
 
== Problem ==
 
Given that<br>
 
Given that<br>
<center><math>\begin{eqnarray*}&(1)& \text{x and y are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\
+
<cmath>\begin{eqnarray*}&(1)& x\text{ and }y\text{ 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}\\
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     &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\
     &(3)& z=|x-y|. \end{eqnarray*}</math></center>  
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     &(3)& z=|x-y|.
 +
\end{eqnarray*}</cmath>
 
How many distinct values of <math>z</math> are possible?
 
How many distinct values of <math>z</math> are possible?
  
 
== Solution ==
 
== Solution ==
We express the numbers as <math>x=100a+10b+c</math> and <math>y=100c+10b+a</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>y=100c+10b+a</math>.  From this, we have  
\end{eqnarray*}</math></center>
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<cmath>\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\
 +
\end{eqnarray*}</cmath>
 
Because <math>a</math> and <math>c</math> are digits, and <math>a</math> is between 1 and 9, there are <math>\boxed{9}</math> possible values.
 
Because <math>a</math> and <math>c</math> are digits, and <math>a</math> is between 1 and 9, there are <math>\boxed{9}</math> possible values.
  

Revision as of 18:54, 10 March 2015

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 $a$ is between 1 and 9, there are $\boxed{9}$ 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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