2007 AIME I Problems/Problem 11

Could someone help with the formatting? I'll post a solution:

$(k-1/2)^2=k^2-k+1/4$ and $(k+1/2)^2=k^2+k+1/4$ Therefore $b(p)=k$ if and only if $p$ is in this range, if and only if $k^2-k<p\leq k^2+k$ . There are $2k$ numbers in this range, so the some of $b(p)$ over this range is $(2k)k=k^2$ . $44<\sqrt{2007}<45$ , so all numbers $1$ to $44$ have their full range. Summing this up we get $2\sum_{k=1}^{44}2k^2=2(44(44+1)(2*44+1)/6)=58740$ . We need only consider the $740$ because we are work modulo $1000$ Now consider the range of numbers such that $b(P)=45$ . These numbers are $1893$ to $2007$ . There are $115$ of them. $115*45=5175$ , and $175+740=935$ , the solution.

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2007 AIME I Problems/Problem 11