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

(Remove extra problem section)
m (Fixed modulo equation)
 
Line 4: Line 4:
 
== 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 \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 \equiv \boxed{640} \pmod {1000}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:33, 18 March 2020

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, $N = 8640 \equiv \boxed{640} \pmod {1000}$.

See also

2010 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS