1996 AMC 8 Problem 14
Attempt Problem 14 of the 1996 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 1996 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
14.
Six different digits from the set are placed in the squares in the figure shown so that the sum of the entries in the vertical column is and the sum of the entries in the horizontal row is The sum of the six digits used is
Answer: B
Small hint:
Three distinct digits summing to must be and
Big hint:
The shared square is counted in both sums, so the total of the six digits is minus the shared digit
Solution:
Three distinct digits from through summing to must be and The row’s other three digits are at least so the shared square (belonging to both the column and the row) is at most Hence the shared digit is
The six digits are then and whose sum is Equivalently,
Thus, the correct answer is B .
Problem 14 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
- 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
- 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