Difference between revisions of "2017 AIME II Problems/Problem 6"
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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. | ||
− | ==Solution== | + | ==Solution 1== |
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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+ | ==Solution 2== | ||
+ | Clearly, the result when <math>n</math> is plugged into the given expression is larger than <math>n</math> itself. Let <math>x</math> be the positive difference between that result and <math>n</math>, so that <math>\sqrt{n^2+85n+2017}=n+x</math>. Squaring both sides and canceling the <math>n^2</math> terms gives <math>85n+2017=2xn+x^2</math>. Combining like terms, <math>(85-2x)n=x^2-2017</math>, so | ||
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+ | <cmath>n=\frac{x^2-2017}{85-2x}.</cmath> | ||
+ | |||
+ | Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>. | ||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=5|num-a=7}} | {{AIME box|year=2017|n=II|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:00, 23 March 2017
Contents
Problem
Find the sum of all positive integers such that is an integer.
Solution 1
Manipulating the given expression, . The expression under the radical must be an square number for the entire expression to be an integer, so . Rearranging, . By difference of squares, . It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, is found to be and . The two values of that satisfy one of the equations are and . Summing these together, the answer is .
Solution 2
Clearly, the result when is plugged into the given expression is larger than itself. Let be the positive difference between that result and , so that . Squaring both sides and canceling the terms gives . Combining like terms, , so
Since is positive, there are two cases, which are simple (luckily). Remembering that is a positive integer, then and are either both positive or both negative. The smallest value for which is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that (from the numerator) and , which means . This only gives two solutions, . Plugging these into the expression for , we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is .
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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