Difference between revisions of "2010 AMC 10A Problems/Problem 9"

Latest revision as of 12:23, 26 May 2020

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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>?
-<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24</math>
-== Solution ==
-<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>.
-== See also ==
-{{AMC10 box|year=2010|num-b=8|num-a=10|ab=A}}
-[[Category:Introductory Algebra Problems]]