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

(Solution)
(Solution)
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<cmath>\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\&=&|99a-99c|\&=&99|a-c|\
 
<cmath>\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\&=&|99a-99c|\&=&99|a-c|\
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
Because <math>a</math> and <math>c</math> are digits, <math>c</math> is greater than 0 (from condition 1), and <math>a</math> is between 1 and 9, there are <math>\boxed{8}</math> possible values.
+
Because <math>a</math> and <math>c</math> are digits, and <math>a</math> and <math>c</math> are both between 1 and 9 (from condition 1), there are <math>\boxed{9}</math> possible values (since all digits except <math>9</math> can be expressed this way).
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:39, 20 October 2016

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 $a$ and $c$ are both between 1 and 9 (from condition 1), there are $\boxed{9}$ possible values (since all digits except $9$ can be expressed this way).

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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