Problem:
Integers a and b are randomly chosen without replacement from the set of integers with absolute value not exceeding 10 . What is the probability that the polynomial x3+ax2+bx+6 has 3 distinct integer roots?
Answer Choices:
A. 2401
B. 2211
C. 1051
D. 841
E. 631
Solution:
Let r,s, and t be the roots of x3+ax2+bx+6. Then
x3+ax2+bx+6=(x−r)(x−s)(x−t)=x3−(r+s+t)x2+(rs+st+tr)x−rst,
so rst=−6,r+s+t=−a, and rs+st+tr=b. The only triples of distinct integers that satisfy the first of these three equations are (6,1,−1),(3,2,−1),(3,−2,1),(−3,2,1), and (−3,−2,−1), together with their permutations. The corresponding values of a and b are (−6,−1),(−4,1),(−2,−5), (0,−7), and (6,11), respectively.
Notice that these ordered pairs are distinct, but in only 4 of them do both a and b have absolute value not exceeding 10 . There are 21⋅20 equally likely choices for a and b, so the required probability is 21⋅204=(C)1051.
The problems on this page are the property of the MAA's American Mathematics Competitions