Problem:
Let P(x) be the unique polynomial of minimal degree with the following properties:
- P(x) has leading coefficient 1,
- 1 is a root of P(x)−1,
- 2 is a root of P(x−2),
- 3 is a root of P(3x), and
- 4 is a root of 4P(x).
The roots of P(x) are integers, with one exception. The root that is not an integer can be written as nm​, where m and n are relatively prime positive integers. What is m+n?
Answer Choices:
A. 41
B. 43
C. 45
D. 47
E. 49
Solution:
The given conditions imply that
P(1)=1,P(2−2)=P(0)=0,P(3⋅3)=P(9)=0,4P(4)=P(4)=0.
The polynomial Q(x)=x(x−4)(x−9) has the required roots, but Q(1)=24î€ =1. Because the leading coefficient of P(x) is 1 , it follows that P(x) must have an additional factor. To minimize the degree of P(x), that factor must be linear, say x−a. Thus P(x)=(x−a)Q(x), and
1=P(1)=(1−a)⋅1⋅(−3)⋅(−8)=24(1−a).
Thus a=2423​, and the requested sum is 23+24=(D)47​.
The problems on this page are the property of the MAA's American Mathematics Competitions