Difference between revisions of "2003 AIME II Problems/Problem 2"
Anonymous03 (talk | contribs) (→Solution) |
Fuzzy growl (talk | contribs) m (→Problem) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let <math>N</math> be the greatest integer multiple of 8, no two whose digits are the same. What is the remainder when <math>N</math> is divided by 1000 | + | Let <math>N</math> be the greatest integer multiple of 8, no two of whose digits are the same. What is the remainder when <math>N</math> is divided by 1000? |
== Solution == | == Solution == |
Revision as of 22:48, 10 March 2010
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 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 |