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

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==Solution 1==
 
==Solution 1==
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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>.
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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 14:03, 22 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 1

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


~heeeeeeheeeee

Solution 2

We let the two digit palindromes be $AA$, $BB$, and $CC$, which sum to $11(A+B+C)$. Now, we can let $A+B+C=k$. This means we are looking for the smallest $k$ such that $11k>100$ and $11k$ is not a palindrome. Thus, we test $10$ for $k$, which works so $11k=110$, meaning that the sum requested is $1+1+0=\boxed{\textbf{(A)}\ 2}$. ~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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