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 | + | 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>, | + | 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 | ||
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+ | == See also == | ||
+ | {{AIME box|year=2010|before=First Problem|num-a=2|n=II}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:54, 28 October 2021
Contents
[hide]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 |
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