Difference between revisions of "2002 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
− | We | + | 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> | |
− | + | Because <math>a</math> and <math>c</math> are digits, and <math>a</math> is between 1 and 9, there are 9 possible values. | |
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== 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 15:12, 9 August 2008
Problem
Given that
&(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 are possible?
Solution
We express the numbers as and . From this, we have
Because and are digits, and is between 1 and 9, there are 9 possible values.
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
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 |