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 3 two digit palindromes 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 <math>11</math>. The least <math>3</math> digit integer that is the sum of <math>3</math> two digit palindromes is a multiple of <math>11</math>. The least <math>3</math> digit multiple of <math>11</math> is <math>110</math>. The sum of the digits of <math>110</math> is <math>1 + 1 + 0 =</math> <math>\boxed{\textbf{(A)}\ 2}</math>. |
Revision as of 15:03, 22 November 2019
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
All the two digit palindromes are multiples of . The least digit integer that is the sum of two digit palindromes is a multiple of . The least digit multiple of is . The sum of the digits of is .
~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 |
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