Difference between revisions of "2010 AIME II Problems/Problem 1"
m (Fixed modulo equation) |
(→Solution) |
||
Line 5: | Line 5: | ||
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 \equiv \boxed{640} \pmod {1000}</math>. | 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>. | ||
+ | |||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=TVlHqIgMEVQ | ||
== See also == | == See also == |
Latest revision as of 11:54, 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.