2017 AMC 8 Problem 12
Attempt Problem 12 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).
12.
The smallest positive integer greater than that leaves a remainder of when divided by and lies between which of the following pairs of numbers?
and
and
and
and
and
Answer: D
Small hint:
Numbers with remainder after division by and are one more than multiples of their least common multiple.
Big hint:
Find the least common multiple of and , then add .
Video solution:
Click to load, then click again to play
Written solution:
If a number leaves a remainder of when divided by and then it is one more than the least common multiple of these numbers. The least common multiple is so the smallest such positive integer is
Thus, D is the correct answer.
Problem 12 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