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

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==Problem 13==
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==Problem==
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A <math>\textit{palindrome}</math> 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 <math>N</math> 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 <math>N</math>?
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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 <math>N</math> 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 <math>N</math>?
  
 
<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==
 
==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 2 two digit integers 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=[A] 2
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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>.
  
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==See Also==
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*There are other sets of 2-digit numbers that satisfy this rule. Some of them are <math>110 = 11+33+66</math> and <math>110 = 22+33+55</math>
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 +
==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,</math> and <math>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 Solution 1 for inspiring me to create this!
 +
 
 +
EarthSaver 15:13, 11 June 2021 (EDT)
 +
 
 +
==Solution 3 (basically a version of the above solutions)==
 +
As stated above, two-digit palindromes can only be two-digit multiples of 11. We can see that if we add anything that are multiples of 11 together, we will again get a multiple of 11. For instance, <math>11+22=33</math>. Since we know this fact and we are finding the smallest value possible, we can start with the first three-digit multiple of 11 which is <math>110</math>. Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), <math>110</math> fits the bill. We can see that the sum of <math>110</math> 's digits is <math>1+1+0 = \boxed{\textbf{(A) }2}</math>.
 +
 
 +
~yeye
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 +
== Video Solution by Pi Academy (Really Easy 👍) ==
 +
 
 +
https://youtu.be/7xlBdxcsP3I?si=sFJGKXnBUUN7Su-C
 +
 
 +
 
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/IgpayYB48C4?si=AbOfamWIMyCRNHTA&t=3984
 +
~Math-X
 +
 
 +
==Video Solution 1==
 +
https://youtu.be/gOZOCFNXMhE ~ The Learning Royal
 +
 
 +
==Video Solution 2==
 +
https://www.youtube.com/watch?v=bOnNFeZs7S8
 +
 
 +
== Video Solution 3==
 +
 
 +
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=PJpDJ23sOJM&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=14
 +
 
 +
==Video Solution 4==
 +
https://youtu.be/WeQuJEQKVdo
 +
 
 +
~savannahsolver
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 +
==Video Solution (CREATIVE ANALYSIS!!!)==
 +
https://youtu.be/FISgn9laDaI
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
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==See also==
 
{{AMC8 box|year=2019|num-b=12|num-a=14}}
 
{{AMC8 box|year=2019|num-b=12|num-a=14}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:20, 23 November 2024

Problem

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

  • There are other sets of 2-digit numbers that satisfy this rule. Some of them are $110 = 11+33+66$ and $110 = 22+33+55$

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!

EarthSaver 15:13, 11 June 2021 (EDT)

Solution 3 (basically a version of the above solutions)

As stated above, two-digit palindromes can only be two-digit multiples of 11. We can see that if we add anything that are multiples of 11 together, we will again get a multiple of 11. For instance, $11+22=33$. Since we know this fact and we are finding the smallest value possible, we can start with the first three-digit multiple of 11 which is $110$. Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), $110$ fits the bill. We can see that the sum of $110$ 's digits is $1+1+0 = \boxed{\textbf{(A) }2}$.

~yeye

Video Solution by Pi Academy (Really Easy 👍)

https://youtu.be/7xlBdxcsP3I?si=sFJGKXnBUUN7Su-C


Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=AbOfamWIMyCRNHTA&t=3984 ~Math-X

Video Solution 1

https://youtu.be/gOZOCFNXMhE ~ The Learning Royal

Video Solution 2

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

Video Solution 3

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

Video Solution 4

https://youtu.be/WeQuJEQKVdo

~savannahsolver

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/FISgn9laDaI

~Education, the Study of Everything

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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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