2023 AMC 12B Problem 14

Attempt Problem 14 of the 2023 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

For how many ordered pairs (a,b)(a,b) of integers does the polynomial x3+ax2+bx+6x^3+ax^2+bx+6 have 33 distinct integer roots?
55
66
88
77
44
Answer: A
Concepts:Vieta’s Formulasfactorsystematic listing
Small hint:
If the roots are integers p,q,r,p,q,r, then pqr=−6pqr=-6
Big hint:
List the sets of three distinct integers with product −6;-6; each set gives one pair (a,b)(a,b)
Solution:
By Vieta, the three distinct integer roots multiply to −6.-6. The sets of three distinct integers with product −6-6 are {1,2,−3},\{1,2,-3\}, {1,−2,3},\{1,-2,3\}, {−1,2,3},\{-1,2,3\}, {−1,−2,−3},\{-1,-2,-3\}, and {1,−1,6}.\{1,-1,6\}. Each set determines a=−(p+q+r)a=-(p+q+r) and b=pq+qr+rp,b=pq+qr+rp, and all five give different pairs, so there are 55 ordered pairs (a,b).(a,b).
Thus, the correct answer is A.
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