1997 AMC 8 Problem 19

Below is the professionally curated solution for Problem 19 of the 1997 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1997 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:telescopingfraction

Difficulty rating: 1340

19.

If the product 32435465ab=9,\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9, what is the sum of aa and b?b?

1111

1313

1717

3535

3737

Solution:

Note that the numerator of each fraction cancels with the denominator of the fraction to its right.

We can then cancel out all these terms to get a final equation of a2=9a=18b=17. \dfrac{a}{2} = 9 \Rightarrow a = 18 \Rightarrow b = 17.

The desired sum is then 18+17=35. 18 + 17 = 35.

Thus, D is the correct answer.

Problem 19 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 · 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