2017 AMC 8 Problem 20
Attempt Problem 20 of the 2017 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 2017 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
20.
An integer between and inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?
Answer: B
Small hint:
Count favorable four-digit numbers directly, then divide by the number of four-digit integers.
Big hint:
Choose the odd units digit first, then choose the nonzero thousands digit, then the remaining two digits.
Video solution:
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Written solution:
Since the number is odd, the last digit is odd, giving possibilities. The thousands digit cannot be zero or the number we already got, so that gives possibilities. Similarly, the hundreds digit has possibilities, and the tens digit has possibilities. This gives a total of making the probability
Thus, B is the correct answer.
Problem 20 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 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
- 1998 AMC 8
- 1999 AMC 8
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