Problem:
The polynomial x3βax2+bxβ2010 has three positive integer zeros. What is the smallest possible value of a?
Answer Choices:
A. 78
B. 88
C. 98
D. 108
E. 118
Solution:
Let the polynomial be (xβr)(xβs)(xβt) with 0<rβ€sβ€t. Then rst=2010=2β
3β
5β
67, and r+s+t=a. If t=67, then rs=30, and r+s is minimized when r=5 and s=6. In that case a=67+5+6=78. If tξ =67, then a>tβ₯2β
67=134, so the minimum value of a is (A)78β.
The problems on this page are the property of the MAA's American Mathematics Competitions