2024 AMC 10A Problem 7

Attempt Problem 7 of the 2024 AMC 10A 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 2024 AMC 10A solutions, or check the answer key.

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

7.

The product of three integers is 60.60. What is the least possible positive sum of the three integers?
22
33
55
66
1313
Answer: B
Concepts:factoroptimizationcasework
Small hint:
A positive product uses either three positive factors or one positive and two negative factors
Big hint:
Two negative factors let the positive factor be large; minimize the positive factor minus the sum of the two magnitudes
Solution:
A positive product comes from three positive integers or one positive and two negative integers. Three positive integers have sum at least 3.3. In the second case write the numbers as −x,−y,z,-x,-y,z, where x,y,zx,y,z are positive and xyz=60.xyz=60. A positive sum requires z>x+y≥2xy,z \gt x+y \ge 2\sqrt{xy}, so 60xy>2xy\frac{60}{xy} \gt 2\sqrt{xy} and hence xy<10.xy \lt 10. The possible products xyxy that divide 6060 are 1,2,3,4,5,6.1,2,3,4,5,6. Checking their factor pairs, the smallest positive value of 60xy−x−y\frac{60}{xy}-x-y is 10−1−6=3.10-1-6=3. Thus (−1)(−6)(10)=60,(-1)(-6)(10)=60, and no positive sum below 33 is possible. Therefore, the answer is B.
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