Problem:
Let m be the largest real solution to the equation
x−33​+x−55​+x−1717​+x−1919​=x2−11x−4
There are positive integers a,b, and c such that m=a+b+c​​. Find a+b+c.
Solution:
Because x−aa​=x−ax​−1, the given equation is equivalent to
x(x−11)=x−3x​+x−17x​+x−5x​+x−19x​.
Thus x=0 is a solution of the equation. If y=x−11, then the rest of the solutions satisfy
y=y+81​+y−81​+y+61​+y−61​=y2−642y​+y2−362y​.
Thus y=0 (that is, x=11 ) is a solution of the equation. If z=y2−50, then the rest of the solutions satisfy
1=z−142​+z+142​
from which it follows that z2−142=4z. Therefore z=±200​+2, and
(x−11)2=y2=z+50=52±200​
and
x=11±52±200​​
implying that m=11+52+200​​. The requested sum is 11+52+200=263​.
The problems on this page are the property of the MAA's American Mathematics Competitions