2010 AMC 12A Problems/Problem 6
- The following problem is from both the 2010 AMC 12A #6 and 2010 AMC 10A #9, so both problems redirect to this page.
Contents
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
Solution 1
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 .
Solution 2
For to be a four-digit number, is in between and . The palindromes in this range are , , , and , so the sum of digits of can be , , , or . Only is an option, and upon checking, is indeed a palindrome.
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=1444
~ pi_is_3.14
Video Solution by the Beauty of Math
https://www.youtube.com/watch?v=P7rGLXp_6es
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.