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 $168+27=\boxed{195}$ .

@@ Line 6: / Line 6: @@
 =See Also=
-{{AIME box|year=2017|n=II|num-b=5|num-a=6}}
+{{AIME box|year=2017|n=II|num-b=5|num-a=7}}
 {{MAA Notice}}

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

Page

Toolbox

Search

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

Revision as of 11:53, 23 March 2017

Problem

Solution

See Also