1990 AMC 8 Problem 11
Attempt Problem 11 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).
11.
The numbers on the faces of this cube are consecutive whole numbers. The sums of the two numbers on each of the three pairs of opposite faces are equal. The sum of the six numbers on this cube is
Answer: E
Small hint:
Five of the faces must be ; the sixth is either or
Big hint:
The three faces meet at one corner, so no two of them are opposite; this rules out one of the two possibilities
Solution:
The six consecutive numbers include , so five of the faces are and the sixth is or .
If the sixth were , equal opposite sums would force the pairs , making and opposite. But the figure shows meeting at one corner, so no two of them are opposite. Hence the sixth number is , with pairs , each summing to .
The total is .
Thus, the correct answer is E .
Problem 11 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
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