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"

 
Line 1: Line 1:
 
Could someone help with the formatting? I'll post a solution:
 
Could someone help with the formatting? I'll post a solution:
  
−
<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.
+
<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.

Revision as of 18:55, 15 March 2007

Could someone help with the formatting? I'll post a 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 $175+740=935$, the solution.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_11&oldid=14045"