Difference between revisions of "2003 AIME II Problems/Problem 2"
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Therefore, the remainder when the number is divided by <math>1000</math> is <math>\boxed{120}</math>. | Therefore, the remainder when the number is divided by <math>1000</math> is <math>\boxed{120}</math>. | ||
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| + | Note: The number is <math>9876543120</math> | ||
== See also == | == See also == | ||
Revision as of 08:00, 11 July 2023
Problem
Let 
be the greatest integer multiple of 8, no two of whose digits are the same. What is the remainder when is divided by 1000?
Solution
We want a number with no digits repeating, so we can only use the digits 
once in constructing our number. To make the greatest number, we want the greatest digit to occupy the leftmost side and the least digit to occupy the rightmost side. Therefore, the last three digits of the greatest number should be an arrangement of the digits 
. Since the number has to be divisible by 8, the integer formed by the arrangement of is also divisible by 8. The only arrangement that works is 
.
Therefore, the remainder when the number is divided by 
is 
.
Note: The number is 
See also
| 2003 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
