Mock AIME 4 2006-2007 Problems/Problem 12

Revision as of 20:34, 27 September 2019 by Someonenumber011 (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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 $a + b$.


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

Invalid username
Login to AoPS