1990 AMC 8 Problem 7
Attempt Problem 7 of the 1990 AMC 8 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 1990 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
7.
When three different numbers from the set are multiplied, the largest possible product is
Answer: C
Small hint:
A product of three numbers is positive only when none or two of them are negative
Big hint:
There are only two positive numbers, so use two negatives and the largest positive
Solution:
For the product of three numbers to be positive, either all three are positive or exactly two are negative. There are only two positive numbers, so we must use two negatives and one positive.
To maximize, take the two most negative numbers and the largest positive: .
Thus, the correct answer is C .
Problem 7 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
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- 1989 AMC 8
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