Difference between revisions of "2002 AIME II Problems/Problem 1"
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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> | + | 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
How many distinct values of 
are possible?
Solution
We express the numbers as 
and 
. From this, we have
Because 
and 
are digits, and and are both between 1 and 9 (from condition 1), there are 
possible values (since all digits except 
can be expressed this way).
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
