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 17:27, 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 |
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