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

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(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>
  
==Solution 1==
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==Solution==
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>.
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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>.
 
 
 
 
~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 02:23, 27 November 2019

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

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 $1+1+0 = \boxed{\textbf{(A) }2}$.

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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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