2007 AIME I Problems/Problem 11

Problem

For each positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$ . For example, $b(6) = 2$ and $b(23) = 5$ . If $S = \Sigma_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000.

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 $175+740=955$ , the solution.

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2007 AIME I Problems/Problem 11

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