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

2007 iTest Problems/Problem 40

Revision as of 13:59, 16 June 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 40 - a lot of work)

Problem

Let $S$ be the sum of all $x$ such that $1\leq x\leq 99$ and $\{x^2\}=\{x\}^2$. Compute $\lfloor S\rfloor$.

Solution

Rewrite $x$ as $\lfloor x \rfloor + \{ x \}$. That results in \[\{ 2 \lfloor x \rfloor \{ x \} + \{ x \}^2 \} = \{ x \}^2\] Note that if $2 \lfloor x \rfloor \{ x \}$ is not an integer, than both sides can not equal each other. Thus, to find all solutions where $1 \le x \le 99$, find the values where $2 \lfloor x \rfloor \{ x \}$ is an integer.

  • The easier case is when $x$ is an integer. If $x$ is an integer, then $\{ x \} = 0$, so both sides equal each other. Therefore, every integer from $1$ to $99$ is a solution.
  • The harder case is when $x$ is not an integer. Let $x = n \frac{a}{b}$, where $n$ is an integer from $1$ to $98$, and $a$ and $b$ are relatively prime integers, where $0 < a < b$. That means $2 \lfloor x \rfloor \{ x \} = 2n \cdot \frac{a}{b}$. Since $a$ and $b$ are relatively prime, $b$ must be a factor of $2n$. That means every mixed number in the form $n \frac{c}{2n}$, where $0 < c < 2n$ works.

With that taken in consideration, the sum $S$ equals \[(1 + 1\frac{1}{2}) + (2 + 2\frac{1}{4} \cdots) \cdots (98 + 98\frac{1}{196} \cdots) + 99\] The arithmetic series sum formula can be used to simplify things further. If the first term is $n$, common difference is \frac{1}{2n}, and last term is $n - \frac{1}{2n}$, the sum of the terms in the series is $\frac{2n(2n+1 - \frac{1}{2n})}{2} = 2n^2 + n - \frac{1}{2}$. Now $S$ equals \[\sum_{n=1}^{98} (2n^2 + n - \frac{1}{2} ) + 99\] \[2 \sum_{n=1}^{98} (n^2) + \sum_{n=1}^{98} (n) - 98 \cdot \frac{1}{2} + 99\] \[2 \cdot \frac{98 \cdot 99 \cdot 197}{6} + \frac{99 \cdot 98}{2} - 49 + 99\] \[49 \cdot 33 \cdot 397 + 50\] \[641999\] Thus, $S = \boxed{641999}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 39 		Followed by:
Problem 41
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_40&oldid=95206"