Problem:
The polynomial x3−2004x2+mx+n has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of n are possible?
Answer Choices:
A. 250,000
B. 250,250
C. 250,500
D. 250,750
E. 251,000
Solution:
Let a denote the zero that is an integer. Because the coefficient of x3 is 1 , there can be no other rational zeros, so the two other zeros must be 2a±r for some irrational number r. The polynomial is then
(x−a)[x−(2a+r)][x−(2a−r)]=x3−2ax2+(45a2−r2)x−a(41a2−r2)
Therefore a=1002 and the polynomial is
x3−2004x2+(5(501)2−r2)x−1002((501)2−r2)
All coefficients are integers if and only if r2 is an integer, and the zeros are positive and distinct if and only if 1≤r2≤5012−1=251,000. Because r cannot be an integer, there are 251,000−500=250,500 possible values of n.
The problems on this page are the property of the MAA's American Mathematics Competitions