2007 AMC 8 Problem 24

Attempt Problem 24 of the 2007 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 2007 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

24.

A bag contains four pieces of paper, each labeled with one of the digits 1,1, 2,2, 33 or 4,4, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?3?

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

Answer: C
Concepts:divisibilitybasic probabilitydigits
Difficulty rating: 1580
Solution:

Recall that a number is divisible by 33 if the sum of its digits is divisible by 33.

The only triples of distinct digits from 1,2,3,41,2,3,4 whose sum is divisible by 33 are (1,2,3)(1,2,3) and (2,3,4)(2,3,4).

This means the constructed number is a multiple of 33 exactly when either 11 or 44 is left in the bag.

Each of these events has probability 14\dfrac{1}{4}, for a total probability of 214=122\cdot\dfrac{1}{4}=\dfrac{1}{2}.

Thus, C is the correct answer.

← Problem 23#23
Full Exam

Problem 24 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 · 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 · 2017 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