2017 AMC 8 Problem 20

Below is the video solution and professionally curated solution for Problem 20 of the 2017 AMC 8, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:basic probabilitymultiplication principle

Difficulty rating: 1550

20.

An integer between 10001000 and 9999,9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

1475 \dfrac{14}{75}

56225 \dfrac{56}{225}

107400 \dfrac{107}{400}

725 \dfrac{7}{25}

925 \dfrac{9}{25}

Video solution:
Solution video thumbnail
Play video

Click to load, then click again to play

Written solution:

Since the number is odd, the last digit is odd, giving 55 possibilities. The thousands digit cannot be zero or the number we already got, so that gives 88 possibilities. Similarly, the hundreds digit has 88 possibilities, and the tens digit has 77 possibilities. This gives a total of 5887=2240,5 \cdot 8 \cdot 8 \cdot 7 = 2240, making the probability 22409000=56225.\dfrac{2240}{9000} = \dfrac{56}{225}.

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 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8