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

(Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
The palindrome formula is to add 110 to the number in order to get the next palindrome, a palindrome needs to be as ABBA . 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>.
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The palindrome formula is to add 110 to the number in order to get the next palindrome, a palindrome needs to be in the form as ABBA . 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}}

Revision as of 15:43, 23 May 2023

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 1

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

Solution 2

The palindrome formula is to add 110 to the number in order to get the next palindrome, a palindrome needs to be in the form as ABBA . We can use this in this case to get 2112. 2*1*1*2=4. Therefore, the answer 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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