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