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

m (Solution)
(Solution)
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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|\\
 
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>
 
\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.
+
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.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|before=First Question|num-a=2}}
 
{{AIME box|year=2002|n=II|before=First Question|num-a=2}}

Revision as of 15:23, 9 August 2008

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 $\boxed{9}$ 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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