1990 AMC 8 Problem 1
Attempt Problem 1 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).
1.
What is the smallest sum of two -digit numbers that can be obtained by placing each of the six digits into one of the six boxes of a sum of two -digit numbers?
Answer: C
Small hint:
A digit in the hundreds place counts times, so put the smallest digits there
Big hint:
Hundreds digits should be and , tens digits and , units digits and
Solution:
To make the sum small, the two smallest digits go in the hundreds places, the next two in the tens places, and the two largest in the units places.
One such arrangement is , and every arrangement of this type gives the same sum.
Thus, the correct answer is C .
Problem 1 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
- 1995 AMC 8
- 1996 AMC 8
- 1997 AMC 8
- 1998 AMC 8
- 1999 AMC 8
- 2000 AMC 8
- 2001 AMC 8
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