Difference between revisions of "2002 AMC 8 Problems/Problem 4"
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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>. | ||
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2002|num-b=3|num-a=5}} | ||
Revision as of 23:59, 30 July 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 
.
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
