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Difference between revisions of "2019 AMC 8 Problems/Problem 13"

(solutions 1&2 identical; for completeness, solution should check that 110 actually works)
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<math>\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6</math>
 
<math>\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6</math>
  
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==Solution==
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==Solution 1==
 
Note that the only positive 2-digit palindromes are multiples of 11, namely <math>11, 22, \ldots, 99</math>. Since <math>N</math> is the sum of 2-digit palindromes, <math>N</math> is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so <math>N=110</math> is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as <math>110=77+22+11</math>. Then <math>N = 110</math>, and the sum of the digits of <math>N</math> is <math>1+1+0 = \boxed{\textbf{(A) }2}</math>.
 
Note that the only positive 2-digit palindromes are multiples of 11, namely <math>11, 22, \ldots, 99</math>. Since <math>N</math> is the sum of 2-digit palindromes, <math>N</math> is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so <math>N=110</math> is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as <math>110=77+22+11</math>. Then <math>N = 110</math>, and the sum of the digits of <math>N</math> is <math>1+1+0 = \boxed{\textbf{(A) }2}</math>.
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==Solution 2==
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Associated video - https://www.youtube.com/watch?v=bOnNFeZs7S8
  
 
==See Also==
 
==See Also==

Revision as of 19:43, 17 July 2020

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 $N$ 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 $N$?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

Solution 1

Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \ldots, 99$. Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$. Then $N = 110$, and the sum of the digits of $N$ is $1+1+0 = \boxed{\textbf{(A) }2}$.

Solution 2

Associated video - https://www.youtube.com/watch?v=bOnNFeZs7S8

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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