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Difference between revisions of "2010 AIME II Problems/Problem 1"

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== Solution ==
 
== Solution ==
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If we include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divisible by <math>3</math>, therefore by <math>36</math> as well.
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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>.
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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>
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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 ==
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https://www.youtube.com/watch?v=TVlHqIgMEVQ
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 11:54, 28 October 2021

Problem

Let $N$ 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 $N$ is divided by $1000$.

Solution

If an integer is divisible by $36$, it must also be divisible by $9$ since $9$ is a factor of $36$. It is a well-known fact that, if $N$ is divisible by $9$, the sum of the digits of $N$ is a multiple of $9$. Hence, if $N$ contains all the even digits, the sum of the digits would be $0 + 2 + 4 + 6 + 8 = 20$, which is not divisible by $9$ and thus $36$. The next logical try would be $8640$, which happens to be divisible by $36$. Thus, $N = 8640 \equiv \boxed{640} \pmod {1000}$.


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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