Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 12"
m |
|||
| Line 4: | Line 4: | ||
{{solution}} | {{solution}} | ||
| − | + | Let <math>f(x)</math> denote the number of partitions that have an even number of even parts of <math>x</math>. Testing a few small values for <math>x</math>, we see that <math>f(1)=0, f(2)=0, f(3)=0, f(4)=1, f(5)=1, f(6)=2, f(7)=2, | |
| − | + | f(8)=4, f(9)=4...</math>. | |
| + | Based on our observations, we now conjecture* that for every integer <math>x\ge 4</math>, <math>f(x)=2^{\lfloor\frac{x-4}{3}\rfloor}</math> So plugging in <math>x=2007</math>, we get <math>f(2007)=2^{1001} \rightarrow 2+1001=1003, 1+3=\boxed{004}</math> | ||
| + | *Insert proof of conjecture here | ||
---- | ---- | ||
Revision as of 08:59, 20 April 2015
Problem
The number of partitions of 2007 that have an even number of even parts can be expressed as 
, where 
and 
are positive integers and is prime. Find the sum of the digits of 
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Let 
denote the number of partitions that have an even number of even parts of 
. Testing a few small values for , we see that 
.
Based on our observations, we now conjecture* that for every integer 
, 
So plugging in 
, we get 
- Insert proof of conjecture here
