Difference between revisions of "2010 AMC 10A Problems/Problem 9"
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It follows that <math>x</math> is <math>969</math>, so the sum of the digits is <math>\boxed{\textbf{(E)}\ 24}</math>. | It follows that <math>x</math> is <math>969</math>, so the sum of the digits is <math>\boxed{\textbf{(E)}\ 24}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/P7rGLXp_6es?t=550 | ||
| + | |||
| + | ~IceMatrix | ||
== See also == | == See also == | ||
Revision as of 05:43, 26 May 2020
Contents
Problem
A palindrome, such as 
, 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 
.
Video Solution
https://youtu.be/P7rGLXp_6es?t=550
~IceMatrix
See also
| 2010 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 10 Problems and Solutions | ||
| 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. 
