Difference between revisions of "2002 AMC 8 Problems/Problem 4"

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==See Also==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=3|num-a=5}}
 
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{{MAA Notice}}

Revision as of 00:41, 5 July 2013

Problem

The year 2002 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 2002 that is a palindrome?

$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$

Solution

The palindrome right after 2002 is 2112. The product of the digits of 2112 is $\boxed{\text{(B)}\ 4}$.

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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