1986 AMC 8 Problem 3

Attempt Problem 3 of the 1986 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 1986 AMC 8 solutions, or check the answer key.

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

3.

The smallest sum one could get by adding three different numbers from the set {7,25,−1,12,−3}\{7, 25, -1, 12, -3\} is
−3-3
−1-1
33
55
2121
Answer: C
Concepts:optimization
Small hint:
To make the sum as small as possible, choose the three smallest numbers
Big hint:
The three smallest numbers are −3,-3, −1,-1, and 77
Solution:
The three smallest numbers in the set are −3,-3, −1,-1, and 7.7. Their sum is −3+(−1)+7=3.-3 + (-1) + 7 = 3.
Thus, the correct answer is C .
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.