Difference between revisions of "2010 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
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>. | 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 \ | + | The next logical try would be <math>8640</math>, which happens to be divisible by <math>36</math>. Thus, <math>N = 8640 \equiv \boxed{640} \pmod {1000}</math>. |
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+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=TVlHqIgMEVQ | ||
== See also == | == See also == |
Latest revision as of 12:54, 28 October 2021
Contents
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,
.
Video Solution
https://www.youtube.com/watch?v=TVlHqIgMEVQ
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.