Difference between revisions of "2003 AIME II Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | + | We want a number with no digits repeating, so we can only use the digits 0-9 once in contructing 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 a arrangement of the digits <math>0,1,2</math>. Since the number has to be divisible by 8, the integer formed by the arrangement of <math>0,1,2</math> has to be divisible by 8 too. The only arrangement that is possible is <math>120</math>. | |
| + | |||
| + | Therefore, the remainder when the number is divided by 1000 is <math>\boxed(120)</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=1|num-a=3}} | {{AIME box|year=2003|n=II|num-b=1|num-a=3}} | ||
Revision as of 13:44, 25 January 2008
Problem
Let 
be the greatest integer multiple of 8, no two 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 0-9 once in contructing 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 a arrangement of the digits 
. Since the number has to be divisible by 8, the integer formed by the arrangement of has to be divisible by 8 too. The only arrangement that is possible is 
.
Therefore, the remainder when the number is divided by 1000 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 | ||
