Difference between revisions of "2010 AMC 12A Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | A <math>\ | + | A <math>\text{palindrome}</math>, such as 83438, is a number that remains the same when its digits are reversed. The numbers <math>x</math> and <math>x+32</math> are three-digit and four-digit palindromes, respectively. What is the sum of the digits of <math>x</math>? |
<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24</math> | <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24</math> | ||
Revision as of 16:27, 17 June 2015
Problem
A 
, such as 83438, is a number that remains the same when its digits are reversed. The numbers 
and 
are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?
Solution
is at most 
, so is at most 
. The minimum value of is 
. However, the only palindrome between and 
is 
, which means that must be .
It follows that is 
, so the sum of the digits is 
.
See also
| 2010 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
