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Difference between revisions of "2007 AIME I Problems/Problem 11"
(sure, I'll fix it up (grrr ... this is the problem I forgot to x2 at the end); btw, the answer is 955)
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Could someone help with the formatting? I'll post a solution:
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== Problem ==
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For each [[positive]] [[integer]] <math>p</math>, let <math>b(p)</math> denote the unique positive integer <math>k</math> such that <math>|k-\sqrt{p}| < \frac{1}{2}</math>.  For example, <math>b(6) = 2</math> and <math>b(23) = 5</math>.  If <math>S = \Sigma_{p=1}^{2007} b(p),</math> find the [[remainder]] when <math>S</math> is divided by 1000.
  
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<math>(k-1/2)^2=k^2-k+1/4</math> and<math>(k+1/2)^2=k^2+k+1/4</math> Therefore <math>b(p)=k</math> if and only if <math>p</math> is in this range, if and only if <math>k^2-k<p\leq k^2+k</math>. There are <math>2k</math> numbers in this range, so the some of <math>b(p)</math> over this range is <math>(2k)k=k^2</math>. <math>44<\sqrt{2007}<45</math>, so all numbers <math>1</math> to <math>44</math> have their full range. Summing this up we get <math>2\sum_{k=1}^{44}2k^2=2(44(44+1)(2*44+1)/6)=58740</math>. We need only consider the <math>740</math> because we are work modulo <math>1000</math> Now consider the range of numbers such that <math>b(P)=45</math>. These numbers are <math>1893</math> to <math>2007</math>. There are <math>115</math> of them. <math>115*45=5175</math>, and <math>175+740=935</math>, the solution.
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== Solution ==
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<math>(k-1/2)^2=k^2-k+1/4</math> and<math>(k+1/2)^2=k^2+k+1/4</math> Therefore <math>b(p)=k</math> if and only if <math>p</math> is in this range, if and only if <math>k^2-k<p\leq k^2+k</math>. There are <math>2k</math> numbers in this range, so the some of <math>b(p)</math> over this range is <math>(2k)k=k^2</math>. <math>44<\sqrt{2007}<45</math>, so all numbers <math>1</math> to <math>44</math> have their full range. Summing this up we get <math>2\sum_{k=1}^{44}2k^2=2(44(44+1)(2*44+1)/6)=58740</math>. We need only consider the <math>740</math> because we are work modulo <math>1000</math> Now consider the range of numbers such that <math>b(P)=45</math>. These numbers are <math>1893</math> to <math>2007</math>. There are <math>115</math> of them. <math>115*45=5175</math>, and <math>175+740=955</math>, the solution.
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== See also ==
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{{AIME box|year=2007|n=I|num-b=10|num-a=12}}
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[[Category:Intermediate Number Theory Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 19:06, 15 March 2007

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.

See also

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
All AIME Problems and Solutions
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