Difference between revisions of "2010 AIME II Problems/Problem 1"
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If we include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divisible by <math>3</math>, therefore by <math>36</math> as well. | If we include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divisible by <math>3</math>, therefore by <math>36</math> as well. | ||
The next logical try would be <math>8640</math>, which happens to be divisible by <math>36</math>. Thus <math>N = 8640 \pmod {1000} = \boxed{640}</math> | The next logical try would be <math>8640</math>, which happens to be divisible by <math>36</math>. Thus <math>N = 8640 \pmod {1000} = \boxed{640}</math> | ||
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| + | == See also == | ||
| + | {{AIME box|year=2010|before=First Problem|num-a=2|n=II}} | ||
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 17:14, 3 April 2010
Problem
Let 
be the greatest integer multiple of 
all of whose digits are even and no two of whose digits are the same. Find the remainder when is divided by 
.
Solution
If we include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divisible by 
, therefore by as well.
The next logical try would be 
, which happens to be divisible by . Thus 
See also
| 2010 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
