Problem:
The polynomial x3−ax2+bx−2010 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 (x−r)(x−s)(x−t) with 0<r≤s≤t. Then rst=2010=2â‹…3â‹…5â‹…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 tî€ =67, then a>t≥2â‹…67=134, so the minimum value of a is (A)78​.
The problems on this page are the property of the MAA's American Mathematics Competitions