Difference between revisions of "2010 AIME II Problems/Problem 1"
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== Problem == | == Problem == | ||
| − | Let <math>N</math> be the greatest integer multiple of 36 all of whose digits are even and no two of whose digits are the same. Find the remainder when <math>N</math> is divided by <math>1000</math>. | + | Let <math>N</math> be the greatest integer multiple of <math>36</math> all of whose digits are even and no two of whose digits are the same. Find the remainder when <math>N</math> is divided by <math>1000</math>. |
== Solution == | == Solution == | ||
If assume to include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divided by <math>3</math> or <math>36</math>. | If assume to include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divided by <math>3</math> or <math>36</math>. | ||
The next logical try would be <math>8640</math>, and this happens to be divisible by <math>36</math>. Thus <math>N = 8640 mod 1000 = \fbox{640}</math> | The next logical try would be <math>8640</math>, and this happens to be divisible by <math>36</math>. Thus <math>N = 8640 mod 1000 = \fbox{640}</math> | ||
Revision as of 15:24, 1 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 assume to include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divided by 
or .
The next logical try would be 
, and this happens to be divisible by . Thus 
