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

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==Solution 2 (variant of Solution 1)==
 
==Solution 2 (variant of Solution 1)==
 
We already know that two-digit palindromes can only be two-digit multiples of 11; which are: <math>11, 22, 33, 44, 55, 66, 77, 88, and 99</math>. Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then we start counting. <math>110 \ldots</math> Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of <math>N</math>’s digits is  <math>1+1+0 = \boxed{\textbf{(A) }2}</math>.
 
We already know that two-digit palindromes can only be two-digit multiples of 11; which are: <math>11, 22, 33, 44, 55, 66, 77, 88, and 99</math>. Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then we start counting. <math>110 \ldots</math> Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of <math>N</math>’s digits is  <math>1+1+0 = \boxed{\textbf{(A) }2}</math>.
 +
 +
Thank you to the writer of <math>Solution 1</math> for inspiring me to create this!
 +
 +
~<math>PrincessBelle</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 12:15, 11 June 2021

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

  • Another set of 2-digit numbers is $110 = 11+33+66$

Solution 2 (variant of Solution 1)

We already know that two-digit palindromes can only be two-digit multiples of 11; which are: $11, 22, 33, 44, 55, 66, 77, 88, and 99$. Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then we start counting. $110 \ldots$ Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of $N$’s digits is $1+1+0 = \boxed{\textbf{(A) }2}$.

Thank you to the writer of $Solution 1$ for inspiring me to create this!

~$PrincessBelle$

Video Solution

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

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=PJpDJ23sOJM&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=14

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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