2016 AMC 8 Problem 24
Attempt Problem 24 of the 2016 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
24.
The digits and are each used once to write a five-digit number The three-digit number is divisible by the three-digit number is divisible by and the three-digit number is divisible by What is
Answer: A
Small hint:
The divisibility by condition determines .
Big hint:
Once is known, list the possible two-digit endings that make divisible by .
Video solution:
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Written solution:
Since is divisible by we know that
Since is divisible by equals either or
If then so cannot be divisible by using the remaining digits. If then again, giving the same obstacle. Thus and then Among the remaining digits, only makes divisible by .
Therefore,
Thus, A is the correct answer.
Problem 24 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
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- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
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