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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)
(→‎Solution: correct solution (avg of 44^2 and 45^2 is -> 1981), fmt)
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== Solution ==
 
== 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.
+
<math>(k- \frac 12)^2=k^2-k+\frac 14</math> and<math>(k+ \frac 12)^2=k^2+k+ \frac 14</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 sum of <math>\displaystyle b(p)</math> over this range is <math>\displaystyle (2k)k=2k^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 with the formula for the sum of the first <math>n</math> squares (<math>\frac{n(n+1)(2n+1)}{6}</math>), we get <math>\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740</math>. We need only consider the <math>740</math> because we are working with modulo <math>1000</math>.
 +
 
 +
Now consider the range of numbers such that <math>\displaystyle b(p)=45</math>. These numbers are <math>\lceil\frac{44^2 + 45^2}{2}\rceil = 1981</math> to <math>2007</math>. There are <math>2007 - 1981 + 1 = 27</math> (1 to be inclusive) of them. <math>27*45=1215</math>, and <math>215+740=955</math>, the solution.
  
 
== See also ==
 
== See also ==

Revision as of 19:13, 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

$(k- \frac 12)^2=k^2-k+\frac 14$ and$(k+ \frac 12)^2=k^2+k+ \frac 14$ 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 sum of $\displaystyle b(p)$ over this range is $\displaystyle (2k)k=2k^2$. $44<\sqrt{2007}<45$, so all numbers $1$ to $44$ have their full range. Summing this up with the formula for the sum of the first $n$ squares ($\frac{n(n+1)(2n+1)}{6}$), we get $\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740$. We need only consider the $740$ because we are working with modulo $1000$.

Now consider the range of numbers such that $\displaystyle b(p)=45$. These numbers are $\lceil\frac{44^2 + 45^2}{2}\rceil = 1981$ to $2007$. There are $2007 - 1981 + 1 = 27$ (1 to be inclusive) of them. $27*45=1215$, and $215+740=955$, 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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