1989 AMC 8 Problem 14
Attempt Problem 14 of the 1989 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 1989 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
14.
When placing each of the digits in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?
Answer: C
Small hint:
Make the three-digit number as small as possible and the two-digit number as large as possible
Big hint:
The largest two-digit number from these digits uses and build the smallest three-digit number from the digits that remain
Solution:
The difference is smallest when the three-digit number (the top) is as small as possible and the two-digit number (subtracted) is as large as possible.
The largest two-digit number is using the digits and The smallest three-digit number from the remaining digits is So the smallest difference is
Thus, the correct answer is C .
Problem 14 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
- 1995 AMC 8
- 1996 AMC 8
- 1997 AMC 8
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- 1999 AMC 8
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