Difference between revisions of "2010 AIME II Problems/Problem 1"
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| − | If | + | If an integer is divisible by <math>36</math>, it must also be divisible by <math>9</math> since <math>9</math> is a factor of <math>36</math>. It is a well-known fact that, if <math>N</math> is divisible by <math>9</math>, the sum of the digits of <math>N</math> is a multiple of <math>9</math>. Hence, if <math>N</math> contains all the even digits, the sum of the digits would be <math>0 + 2 + 4 + 6 + 8 = 20</math>, which is not divisible by <math>9</math> and thus <math>36</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> | 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> | ||
Revision as of 18:53, 29 August 2015
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 an integer is divisible by , it must also be divisible by 
since is a factor of . It is a well-known fact that, if is divisible by , the sum of the digits of is a multiple of . Hence, if contains all the even digits, the sum of the digits would be 
, which is not divisible by and thus .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
