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Difference between revisions of "2007 iTest Problems/Problem 40"

(Solution to Problem 40 - a lot of work)
m (→‎Solution)
 
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With that taken in consideration, the sum <math>S</math> equals
 
With that taken in consideration, the sum <math>S</math> equals
 
<cmath>(1 + 1\frac{1}{2}) + (2 + 2\frac{1}{4} \cdots) \cdots (98 + 98\frac{1}{196} \cdots) + 99</cmath>
 
<cmath>(1 + 1\frac{1}{2}) + (2 + 2\frac{1}{4} \cdots) \cdots (98 + 98\frac{1}{196} \cdots) + 99</cmath>
−
The arithmetic series sum formula can be used to simplify things further.  If the first term is <math>n</math>, common difference is \frac{1}{2n}, and last term is <math>n - \frac{1}{2n}</math>, the sum of the terms in the series is <math>\frac{2n(2n+1 - \frac{1}{2n})}{2} = 2n^2 + n - \frac{1}{2}</math>.  Now <math>S</math> equals
+
The arithmetic series sum formula can be used to simplify things further.  If the first term is <math>n</math>, common difference is <math>\frac{1}{2n}</math>, and last term is <math>n - \frac{1}{2n}</math>, the sum of the terms in the series is <math>\frac{2n(2n+1 - \frac{1}{2n})}{2} = 2n^2 + n - \frac{1}{2}</math>.  Now <math>S</math> equals
 
<cmath>\sum_{n=1}^{98} (2n^2 + n - \frac{1}{2} ) + 99</cmath>
 
<cmath>\sum_{n=1}^{98} (2n^2 + n - \frac{1}{2} ) + 99</cmath>
 
<cmath>2 \sum_{n=1}^{98} (n^2) + \sum_{n=1}^{98} (n) - 98 \cdot \frac{1}{2} + 99</cmath>
 
<cmath>2 \sum_{n=1}^{98} (n^2) + \sum_{n=1}^{98} (n) - 98 \cdot \frac{1}{2} + 99</cmath>
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<cmath>49 \cdot 33 \cdot 397 + 50</cmath>
 
<cmath>49 \cdot 33 \cdot 397 + 50</cmath>
 
<cmath>641999</cmath>
 
<cmath>641999</cmath>
−
Thus, <math>S = \boxed{641999}</math>.
+
Thus, <math>\lfloor S \rfloor = \boxed{641999}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 19:17, 16 June 2018

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, $\lfloor S \rfloor = \boxed{641999}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 39 		Followed by:
Problem 41
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