Difference between revisions of "2003 AIME II Problems/Problem 2"
Fuzzy growl (talk | contribs) m (→Solution) |
|||
| Line 9: | Line 9: | ||
== 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}} | ||
| + | {{MAA Notice}} | ||
Revision as of 19:38, 4 July 2013
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 0-9 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 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
