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

(Created page with "== Problem == Let <math>S</math> be the sum of all <math>x</math> such that <math>1\leq x\leq 99</math> and <math>\{x^2\}=\{x\}^2</math>. Compute <math>\lfloor S\rfloor</math>. ...")
 
(Solution to Problem 40 - a lot of work)
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== Solution ==
 
== Solution ==
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Rewrite <math>x</math> as <math>\lfloor x \rfloor + \{ x \}</math>.  That results in
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<cmath>\{ 2 \lfloor x \rfloor \{ x \} + \{ x \}^2 \} = \{ x \}^2</cmath>
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Note that if <math>2 \lfloor x \rfloor \{ x \}</math> is not an integer, than both sides can not equal each other.  Thus, to find all solutions where <math>1 \le x \le 99</math>, find the values where <math>2 \lfloor x \rfloor \{ x \}</math> is an integer.
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 +
* The easier case is when <math>x</math> is an integer.  If <math>x</math> is an integer, then <math>\{ x \} = 0</math>, so both sides equal each other.  Therefore, every integer from <math>1</math> to <math>99</math> is a solution.
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* The harder case is when <math>x</math> is not an integer.  Let <math>x = n \frac{a}{b}</math>, where <math>n</math> is an integer from <math>1</math> to <math>98</math>, and <math>a</math> and <math>b</math> are [[relatively prime]] integers, where <math>0 < a < b</math>.  That means <math>2 \lfloor x \rfloor \{ x \} = 2n \cdot \frac{a}{b}</math>.  Since <math>a</math> and <math>b</math> are relatively prime, <math>b</math> must be a factor of <math>2n</math>.  That means every mixed number in the form <math>n \frac{c}{2n}</math>, where <math>0 < c < 2n</math> works.
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With that taken in consideration, the sum <math>S</math> equals
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<cmath>(1 + 1\frac{1}{2}) + (2 + 2\frac{1}{4} \cdots) \cdots (98 + 98\frac{1}{196} \cdots) + 99</cmath>
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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
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<cmath>\sum_{n=1}^{98} (2n^2 + n - \frac{1}{2} ) + 99</cmath>
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<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>2 \cdot \frac{98 \cdot 99 \cdot 197}{6} + \frac{99 \cdot 98}{2} - 49 + 99</cmath>
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<cmath>49 \cdot 33 \cdot 397 + 50</cmath>
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<cmath>641999</cmath>
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Thus, <math>S = \boxed{641999}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=39|num-a=41}}
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 13:59, 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, $S = \boxed{641999}$.

See Also

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