Mock AIME 4 2006-2007 Problems/Problem 12

Problem

The number of partitions of 2007 that have an even number of even parts can be expressed as $a^b$ , where $a$ and $b$ are positive integers and $a$ is prime. Find the sum of the digits of $a + b$ .

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Let $f(x)$ denote the number of partitions that have an even number of even parts of $x$ . Testing a few small values for $x$ , we see that $f(1)=0, f(2)=0, f(3)=0, f(4)=1, f(5)=1, f(6)=2, f(7)=2, f(8)=4, f(9)=4...$ . Based on our observations, we now conjecture* that for every integer $x\ge 4$ , $f(x)=2^{\lfloor\frac{x-4}{3}\rfloor}$ So plugging in $x=2007$ , we get $f(2007)=2^{1001} \rightarrow 2+1001=1003, 1+3=\boxed{004}$

Insert proof of conjecture here

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Mock AIME 4 2006-2007 Problems/Problem 12

Problem

Solution