Difference between revisions of "2019 AMC 8 Problems/Problem 13"
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*Another set of 2-digit numbers is <math>110 = 11+33+66</math> | *Another set of 2-digit numbers is <math>110 = 11+33+66</math> | ||
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+ | ==Solution 2 (variant of Solution 1)== | ||
+ | We already know that two-digit palindromes can only be two-digit multiples of 11; which are: <math>11, 22, 33, 44, 55, 66, 77, 88, and 99</math>. Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then we start counting. <math>110 \ldots</math> Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of <math>N</math>’s digits is <math>1+1+0 = \boxed{\textbf{(A) }2}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:13, 11 June 2021
Contents
Problem 13
A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of ?
Solution 1
Note that the only positive 2-digit palindromes are multiples of 11, namely . Since is the sum of 2-digit palindromes, is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as . Then , and the sum of the digits of is .
- Another set of 2-digit numbers is
Solution 2 (variant of Solution 1)
We already know that two-digit palindromes can only be two-digit multiples of 11; which are: . Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then we start counting. Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of ’s digits is .
Video Solution
https://www.youtube.com/watch?v=bOnNFeZs7S8
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=PJpDJ23sOJM&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=14
See also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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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