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

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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 <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>?
 
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>?

Revision as of 13:19, 26 May 2020

The following problem is from both the 2010 AMC 12A #6 and 2010 AMC 10A #9, so both problems redirect to this page.

Problem

A palindrome, such as $83438$, is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$?

$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$

Solution

$x$ is at most $999$, so $x+32$ is at most $1031$. The minimum value of $x+32$ is $1000$. However, the only palindrome between $1000$ and $1032$ is $1001$, which means that $x+32$ must be $1001$.

It follows that $x$ is $969$, so the sum of the digits is $\boxed{\textbf{(E)}\ 24}$.

Video Solution

https://youtu.be/P7rGLXp_6es?t=550

~IceMatrix

See also

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
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

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