Let f(n)=n3β5n2+2n+8, and let g(n)=n3β6n2+5n+12. What is the sum of all integer values of n for which g(n)f(n)β is also an integer?
Answer Choices:
A. 2
B. 3
C. 4
D. 5
E. 6
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Let
f(n)=n3β5n2+2n+8g(n)=n3β6n2+5n+12.
Using the Remainder Theorem,
f(β1)=f(2)=f(4)=0g(β1)=g(3)=g(4)=0ββf(n)=(n+1)(nβ2)(nβ4),βg(n)=(n+1)(nβ3)(nβ4).β
Hence,
g(n)f(n)β=(n+1)(nβ3)(nβ4)(n+1)(nβ2)(nβ4)β=nβ3nβ2β,
with the restrictions
nξ =β1,3,4
(since these make the denominator of the original fraction zero). Now,
g(n)f(n)β=nβ3nβ2β=nβ3(nβ3)+1β=1+nβ31β.
For g(n)f(n)β to be an integer, we must have
nβ31ββZ.
Thus nβ3 must divide 1, so
nβ3=Β±1βn=4 or n=2.
But n=4 is not allowed (it makes g(4)=0), so the only integer value of n satisfying the condition is n=2. Therefore, the sum of all integer values of n for the given condition is (A) 2β.
The problems on this page are the property of the MAA's American Mathematics Competitions