Difference between revisions of "2019 AMC 8 Problems/Problem 13"
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==Problem 13== | ==Problem 13== | ||
− | A | + | 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== | ||
− | + | 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>. | |
− | |||
− | ==See | + | *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> |
+ | |||
+ | ==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 | ||
+ | |||
+ | ==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(Problem 1 to 25 Full Solution)== | ||
+ | https://youtu.be/Xm4ZGND9WoY | ||
+ | |||
+ | ~Hayabusa1 | ||
+ | |||
+ | ==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 15:28, 30 December 2023
Contents
- 1 Problem 13
- 2 Solution 1
- 3 Solution 2 (variant of Solution 1)
- 4 Solution 3 (basically a version of the above solutions)
- 5 Video Solution by Math-X (First fully understand the problem!!!)
- 6 Video Solution 1
- 7 Video Solution 2
- 8 Video Solution 3
- 9 Video Solution 4
- 10 Video Solution (CREATIVE ANALYSIS!!!)
- 11 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 12 See also
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 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 ?
Solution 1
Note that the only positive 2-digit palindromes are multiples of 11, namely . Since is the sum of 2-digit palindromes, is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as . Then, , and the sum of the digits of is .
- There are other sets of 2-digit numbers that satisfy this rule. Some of them are and
Solution 2 (variant of Solution 1)
We already know that two-digit palindromes can only be two-digit multiples of 11; which are: and . Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then, we start counting. 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 ’s digits is .
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, . 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 . Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), fits the bill. We can see that the sum of 's digits is .
~yeye
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
~savannahsolver
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.