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Difference between revisions of "2017 AIME II Problems/Problem 6"

(Created page with "<math>\textbf{Problem 6}</math> Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer. <math>\textbf{Problem 6 Solutio...")
 
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<math>\textbf{Problem 6}</math>
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==Problem==
 
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.
 
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.
  
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<math>\textbf{Problem 6 Solution}</math>
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==Solution==
 
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.
 
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.
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=See Also=
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{{AIME box|year=2017|n=II|num-b=5|num-a=6}}
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{{MAA Notice}}

Revision as of 11:53, 23 March 2017

Problem

Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer.

Solution

Manipulating the given expression, $\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}$. The expression under the radical must be an square number for the entire expression to be an integer, so $(2n+85)^2+843=s^2$. Rearranging, $s^2-(2n+85)^2=843$. By difference of squares, $(s-(2n+85))(s+(2n+85))=1\times843=3\times281$. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, $2n+85$ is found to be $421$ and $139$. The two values of $n$ that satisfy one of the equations are $168$ and $27$. Summing these together, the answer is $168+27=\boxed{195}$.

See Also

2017 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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