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 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.

Video Solution

https://youtu.be/ZhAZ1oPe5Ds?t=1444

~ pi_is_3.14

Video Solution by the Beauty of Math

https://www.youtube.com/watch?v=P7rGLXp_6es

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search