Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
Line 6: Line 6:
  
 
Now consider the range of numbers such that <math>b(p)=45</math>. These numbers are <math>\left\lceil\frac{44^2 + 45^2}{2}\right\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=
 
Now consider the range of numbers such that <math>b(p)=45</math>. These numbers are <math>\left\lceil\frac{44^2 + 45^2}{2}\right\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=
−
boxed{955}</math>, the solution.
+
\boxed{955}</math>, the solution.
  
 
== See also ==
 
== See also ==

Revision as of 04:20, 19 December 2017

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

$\left(k- \frac 12\right)^2=k^2-k+\frac 14$ 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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_11&oldid=88980"