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

Revision as of 14:12, 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 $100c+10b+c$ . 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 count the number of three-letter and three-digit palindromes, then subtract the number of license plates containing both types of palindrome.
+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|\\
+\end{eqnarray*}</math></center>
-There are <math>10^3\cdot 26^2</math> letter palindromes, <math>10^2\cdot 26^3</math> digit palindromes, and <math>10^2\cdot26^2</math> palindromes that contain both letters and digits.
+Because <math>a</math> and <math>c</math> are digits, and <math>a</math> is between 1 and 9, there are 9 possible values.
-Since there are <math>10^3\cdot26^3</math> possible plates, the probability desired is <math>\frac{10^2\cdot26^2(10+26-1)}{10^2\cdot26^2\cdot 260}=\frac{35}{260}=\frac{7}{52}</math>. Thus <math>m+n=059</math>.
 == See also ==
 {{AIME box|year=2002|n=II|before=First Question|num-a=2}}

2002 AIME II (Problems • Answer Key • Resources)
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