Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (variant of Solution 1))
Line 15: Line 15:
  
 
EarthSaver 15:13, 11 June 2021 (EDT)
 
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
 +
  
 
==Video Solution==
 
==Video Solution==

Revision as of 09:25, 14 August 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!

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_13&oldid=160236"