Difference between revisions of "2002 AIME II Problems/Problem 1"
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== Problem == | == Problem == | ||
Given that<br> | Given that<br> | ||
| − | < | + | <cmath>\begin{eqnarray*}&(1)& x\text{ and }y\text{ are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\ |
| − | &(2)& \text{ | + | &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\ |
| − | &(3)& z=|x-y|. \end{eqnarray*}</ | + | &(3)& z=|x-y|. |
| + | \end{eqnarray*}</cmath> | ||
How many distinct values of <math>z</math> are possible? | How many distinct values of <math>z</math> are possible? | ||
== Solution == | == Solution == | ||
| − | We express the numbers as <math>x=100a+10b+c</math> and <math>y=100c+10b+a</math>. From this, we have < | + | We express the numbers as <math>x=100a+10b+c</math> and <math>y=100c+10b+a</math>. From this, we have |
| − | \end{eqnarray*}</ | + | <cmath>\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\ |
| + | \end{eqnarray*}</cmath> | ||
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. | 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. | ||
Revision as of 17:54, 10 March 2015
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 is between 1 and 9, there are 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
