2002 AMC 8 Problem 4
Attempt Problem 4 of the 2002 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 2002 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
4.
The year is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after that is a palindrome?
Answer: B
Small hint:
The next palindrome after still starts and ends with
Big hint:
The middle two digits must increase to the smallest matching pair.
Solution:
We don’t want to increase the thousands digit, so we can keep that as
This means that we have to increase the tens and hundreds digits to to yield the next palindrome of The product of its digits is
Thus, B is the correct answer.
Problem 4 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
- 2003 AMC 8
- 2004 AMC 8
- 2005 AMC 8
- 2006 AMC 8
- 2007 AMC 8
- 2008 AMC 8
- 2009 AMC 8
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- 2011 AMC 8
- 2012 AMC 8
- 2013 AMC 8
- 2014 AMC 8
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- 2020 AMC 8
- 2022 AMC 8
- 2023 AMC 8
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- 2025 AMC 8
- 2026 AMC 8