Difference between revisions of "2010 AMC 12A Problems/Problem 6"
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| + | {{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #6]] and [[2010 AMC 10A Problems|2010 AMC 10A #9]]}} | ||
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== Problem == | == Problem == | ||
A <math>\text{palindrome}</math>, such as 83438, is a number that remains the same when its digits are reversed. The numbers <math>x</math> and <math>x+32</math> are three-digit and four-digit palindromes, respectively. What is the sum of the digits of <math>x</math>? | A <math>\text{palindrome}</math>, such as 83438, is a number that remains the same when its digits are reversed. The numbers <math>x</math> and <math>x+32</math> are three-digit and four-digit palindromes, respectively. What is the sum of the digits of <math>x</math>? | ||
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===Solution 2=== | ===Solution 2=== | ||
For <math>x+32</math> to be a four-digit number, <math>x</math> is in between <math>968</math> and <math>999</math>. The palindromes in this range are <math>969</math>, <math>979</math>, <math>989</math>, and <math>999</math>, so the sum of digits of <math>x</math> can be <math>24</math>, <math>25</math>, <math>26</math>, or <math>27</math>. Only <math>\boxed{\textbf{(E)}\ 24}</math> is an option, and upon checking, <math>x+32=1001</math> is indeed a palindrome. | For <math>x+32</math> to be a four-digit number, <math>x</math> is in between <math>968</math> and <math>999</math>. The palindromes in this range are <math>969</math>, <math>979</math>, <math>989</math>, and <math>999</math>, so the sum of digits of <math>x</math> can be <math>24</math>, <math>25</math>, <math>26</math>, or <math>27</math>. Only <math>\boxed{\textbf{(E)}\ 24}</math> is an option, and upon checking, <math>x+32=1001</math> is indeed a palindrome. | ||
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| + | == Video Solution == | ||
| + | https://youtu.be/ZhAZ1oPe5Ds?t=1444 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
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| + | ==Video Solution by the Beauty of Math== | ||
| + | https://www.youtube.com/watch?v=P7rGLXp_6es | ||
== See also == | == See also == | ||
Revision as of 22:14, 17 January 2021
- The following problem is from both the 2010 AMC 12A #6 and 2010 AMC 10A #9, so both problems redirect to this page.
Contents
Problem
A 
, such as 83438, is a number that remains the same when its digits are reversed. The numbers 
and 
are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?
Solution
Solution 1
is at most 
, so is at most 
. The minimum value of is 
. However, the only palindrome between and 
is 
, which means that must be .
It follows that is 
, so the sum of the digits is 
.
Solution 2
For to be a four-digit number, is in between 
and . The palindromes in this range are , 
, 
, and , so the sum of digits of can be 
, 
, 
, or 
. Only is an option, and upon checking, 
is indeed a palindrome.
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=1444
~ pi_is_3.14
Video Solution by the Beauty of Math
https://www.youtube.com/watch?v=P7rGLXp_6es
See also
| 2010 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
