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

(Redirected page to 2010 AMC 12A Problems/Problem 6)
(Tag: New redirect)
 
(4 intermediate revisions by 2 users not shown)
Line 1: Line 1:
== 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]]
 
{{MAA Notice}}
 

Latest revision as of 12:23, 26 May 2020