# Difference between revisions of "2010 AMC 12A Problems/Problem 6"

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 $\text{palindrome}$, 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

### Solution 1

$x$ 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 $\boxed{\textbf{(E)}\ 24}$.

### Solution 2

For $x+32$ to be a four-digit number, $x$ is in between $968$ and $999$. The palindromes in this range are $969$, $979$, $989$, and $999$, so the sum of the digits of $x$ can be $24$, $25$, $26$, or $27$. Only $\boxed{\textbf{(E)}\ 24}$ is an option, and upon checking, $x+32=1001$ is indeed a palindrome.

### Solution 3

Since we know $x+32$ to be $1 a a 1$ and the only palindrome that works is $0 = a$, that means $x+32 = 1001$, and so $x = 1001 - 32 = 969$. So, $9$ + $6$ + $9$ = $\boxed{\textbf{(E)}\ 24}$. ~songmath20

## Video Solution

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