Difference between revisions of "2002 AMC 8 Problems/Problem 4"
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| + | ==Problem 4== | ||
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| + | 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? | ||
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| + | <math>\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25</math> | ||
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| + | ==Solution== | ||
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The palindrome right after 2002 is 2112. The product of the digits of 2112 is <math>\boxed{4}</math>. | The palindrome right after 2002 is 2112. The product of the digits of 2112 is <math>\boxed{4}</math>. | ||
Revision as of 13:15, 17 June 2011
Problem 4
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?
Solution
The palindrome right after 2002 is 2112. The product of the digits of 2112 is 
.
