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Difference between revisions of "2007 AIME I Problems/Problem 11"

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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 = \sum_{p=1}^{2007} b(p),</math> find the [[remainder]] when <math>S</math> is divided by 1000.
 
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 = \sum_{p=1}^{2007} b(p),</math> find the [[remainder]] when <math>S</math> is divided by 1000.
  
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== Solution ==
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== Solution 1 ==
 
<math>\left(k- \frac 12\right)^2=k^2-k+\frac 14</math> and <math>\left(k+ \frac 12\right)^2=k^2+k+ \frac 14</math>. Therefore <math>b(p)=k</math> if and only if <math>p</math> is in this range, or <math>k^2-k<p\leq k^2+k</math>. There are <math>2k</math> numbers in this range, so the sum of <math>b(p)</math> over this range is <math>(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>.
 
<math>\left(k- \frac 12\right)^2=k^2-k+\frac 14</math> and <math>\left(k+ \frac 12\right)^2=k^2+k+ \frac 14</math>. Therefore <math>b(p)=k</math> if and only if <math>p</math> is in this range, or <math>k^2-k<p\leq k^2+k</math>. There are <math>2k</math> numbers in this range, so the sum of <math>b(p)</math> over this range is <math>(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>.
  

Latest revision as of 19:02, 30 December 2023

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 $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000.

Solution 1

$\left(k- \frac 12\right)^2=k^2-k+\frac 14$ and $\left(k+ \frac 12\right)^2=k^2+k+ \frac 14$. Therefore $b(p)=k$ if and only if $p$ is in this range, or $k^2-k<p\leq k^2+k$. There are $2k$ numbers in this range, so the sum of $b(p)$ over this range is $(2k)k=2k^2$. $44<\sqrt{2007}<45$, 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 $b(p)=45$. These numbers are $\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981$ to $2007$. There are $2007 - 1981 + 1 = 27$ (1 to be inclusive) of them. $27*45=1215$, and $215+740= \boxed{955}$, the answer.

Solution 2

Let $p$ be in the range of $a^2 \le p < (a+1)^2$. Then, we need to find the point where the value of $b(p)$ flips from $k$ to $k+1$. This will happen when $p$ exceeds $(a+\frac{1}{2})^2$ or $a(a+1)+\frac{1}{4}$. Thus, if $a^2 \le p \le a(a+1)$ then $b(p)=a$. For $a(a+1) < p < (a+1)^2$, then $b(p)=a+1$. There are $a+1$ terms in the first set of $p$, and $a$ terms in the second set. Thus, the sum of $b(p)$ from $a^2 \le p <(a+1)^2$ is $2a(a+1)$ or $4\cdot\binom{a+1}{2}$. For the time being, consider that $S = \sum_{p=1}^{44^2-1} b(p)$. Then, the sum of the values of $b(p)$ is $4\binom{2}{2}+4\binom{3}{2}+\cdots +4\binom{44}{2}=4\left(\binom{2}{2}+\binom{3}{2}+\cdots +\binom{44}{2}\right)$. We can collapse this to $4\binom{45}{3}=56760$. Now, we have to consider $p$ from $44^2 \le p < 2007$. Considering $p$ from just $44^2 \le p \le 1980$, we see that all of these values have $b(p)=44$. Because there are $45$ values of $p$ in that range, the sum of $b(p)$ in that range is $45\cdot44=1980$. Adding this to $56760$ we get $58740$ or $740$ mod $1000$. Now, take the range $1980 < p \le 2007$. There are $27$ values of $p$ in this range, and each has $b(p)=45$. Thus, that contributes $27*45=1215$ or $215$ to the sum. Finally, adding $740$ and $215$ we get $740+215=\boxed{955}$.

~firebolt360

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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