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

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==Problem 13==
 
==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 <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==
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>.
  
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*Another set of 2-digit numbers is <math>110 = 11+33+66</math>
  
~heeeeeeheeeee
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==Solution 2 (variant of Solution 1)==
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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>.
  
==Solution 2==
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Thank you to the writer of Solution 1 for inspiring me to create this!
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==
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EarthSaver 15:13, 11 June 2021 (EDT)
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==Solution 3 (basically a version of the above solutions)==
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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>.
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~yeye
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==Solution Explained==
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https://youtu.be/gOZOCFNXMhE ~ The Learning Royal
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 +
==Video Solution==
 +
https://www.youtube.com/watch?v=bOnNFeZs7S8
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== Video Solution ==
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Solution detailing how to solve the problem: https://www.youtube.com/watch?v=PJpDJ23sOJM&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=14
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==Video Solution==
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https://youtu.be/WeQuJEQKVdo
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~savannahsolver
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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}}

Revision as of 11:33, 26 January 2022

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!

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


Solution Explained

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

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

Video Solution

https://youtu.be/WeQuJEQKVdo

~savannahsolver

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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