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| − | == Problem ==
| + | #REDIRECT [[2010_AMC_12A_Problems/Problem_6]] |
| − | A <i>palindrome</i>, 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>?
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| − | <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24</math>
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| − | == Solution ==
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| − | <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>.
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| − | It follows that <math>x</math> is <math>969</math>, so the sum of the digits is <math>\boxed{\textbf{(E)}\ 24}</math>.
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| − | ==Video Solution==
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| − | https://youtu.be/P7rGLXp_6es?t=550
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| − | ~IceMatrix
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| − | == See also ==
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| − | {{AMC10 box|year=2010|num-b=8|num-a=10|ab=A}}
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| − | {{AMC12 box|year=2010|num-b=5|num-a=7|ab=A}}
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| − | [[Category:Introductory Algebra Problems]] | |
| − | {{MAA Notice}}
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