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{\text{(B)}\ 4}</math>. | The palindrome right after 2002 is 2112. The product of the digits of 2112 is <math>\boxed{\text{(B)}\ 4}</math>. | ||
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| + | == Solution 2 == | ||
| + | The palindrome formula is to add 110 to the number in order to get the next palindrome. We can use this in this case to get 2112. 2*1*1*2=4. Therefore, the answer is <math>\boxed{\text{(B)}\ 4}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=3|num-a=5}} | {{AMC8 box|year=2002|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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Revision as of 18:20, 17 January 2023
Contents
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?
Solution 1
The palindrome right after 2002 is 2112. The product of the digits of 2112 is 
.
Solution 2
The palindrome formula is to add 110 to the number in order to get the next palindrome. We can use this in this case to get 2112. 2*1*1*2=4. Therefore, the answer 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
