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 <math>83438</math>, 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>? |
<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24</math> | <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24</math> | ||
== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
<math>x</math> is at most <math>999</math>, so <math>x+32</math> is at most <math>1031</math>. The minimum value of <math>x+32</math> is <math>1000</math>. However, the only palindrome between <math>1000</math> and <math>1032</math> is <math>1001</math>, which means that <math>x+32</math> must be <math>1001</math>. | <math>x</math> is at most <math>999</math>, so <math>x+32</math> is at most <math>1031</math>. The minimum value of <math>x+32</math> is <math>1000</math>. However, the only palindrome between <math>1000</math> and <math>1032</math> is <math>1001</math>, which means that <math>x+32</math> must be <math>1001</math>. | ||
It follows that <math>x</math> is <math>969</math>, so the sum of the digits is <math>\boxed{\textbf{(E)}\ 24}</math>. | It follows that <math>x</math> is <math>969</math>, so the sum of the digits is <math>\boxed{\textbf{(E)}\ 24}</math>. | ||
+ | |||
+ | ===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 the 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. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Since we know <math>x+32</math> to be <math>1 a a 1</math> and the only palindrome that works is <math>0 = a</math>, that means <math>x+32 = 1001</math>, and so <math>x = 1001 - 32 = 969</math>. So, <math>9</math> + <math>6</math> + <math>9</math> = <math>\boxed{\textbf{(E)}\ 24}</math>. | ||
+ | ~songmath20 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/IQj27LEQF4Y | ||
+ | |||
+ | ~Education, the Study of Everything | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=5|num-a=7|ab=A}} | {{AMC12 box|year=2010|num-b=5|num-a=7|ab=A}} | ||
+ | |||
+ | {{AMC10 box|year=2010|num-b=8|num-a=10|ab=A}} | ||
+ | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:45, 27 October 2022
- 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 , 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 the digits of can be , , , or . Only is an option, and upon checking, is indeed a palindrome.
Solution 3
Since we know to be and the only palindrome that works is , that means , and so . So, + + = . ~songmath20
Video Solution
~Education, the Study of Everything
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 |
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.