Difference between revisions of "2019 AMC 8 Problems/Problem 13"
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All the two digit palindromes are multiples of 11. The least 3 digit integer that is the sum of 2 two digit integers is a multiple of 11. The least 3 digit multiple of 11 is 110. The sum of the digits of 110 is 1 + 1 + 0 = <math>\boxed{\textbf{(A)}\ 2}</math>. | All the two digit palindromes are multiples of 11. The least 3 digit integer that is the sum of 2 two digit integers is a multiple of 11. The least 3 digit multiple of 11 is 110. The sum of the digits of 110 is 1 + 1 + 0 = <math>\boxed{\textbf{(A)}\ 2}</math>. | ||
~heeeeeeheeeee | ~heeeeeeheeeee | ||
| + | ==Solution 2== | ||
| + | We let the two digit palindromes be <math>AA</math>, <math>BB</math>, and <math>CC</math>, which sum to <math>11(A+B+C)</math>. Now, we can let <math>A+B+C=k</math>. This means we are looking for the smallest <math>k</math> such that <math>11k>100</math> and <math>11k</math> is not a palindrome. Thus, we test <math>10</math> for <math>k</math>, which works so <math>11k=110</math>, meaning that the sum requested is <math>1+1+0=\boxed{\textbf{(A)}\ 2}</math>. | ||
| + | ~smartninja2000 | ||
==See Also== | ==See Also== | ||
Revision as of 01:59, 21 November 2019
Contents
Problem 13
A 
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
All the two digit palindromes are multiples of 11. The least 3 digit integer that is the sum of 2 two digit integers is a multiple of 11. The least 3 digit multiple of 11 is 110. The sum of the digits of 110 is 1 + 1 + 0 = 
.
~heeeeeeheeeee
Solution 2
We let the two digit palindromes be 
, 
, and 
, which sum to 
. Now, we can let 
. This means we are looking for the smallest 
such that 
and 
is not a palindrome. Thus, we test 
for , which works so 
, meaning that the sum requested is 
.
~smartninja2000
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
