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

Latest revision as of 03:51, 21 January 2023

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 by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/IQj27LEQF4Y

~Education, the Study of Everything

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

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

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #6]] and [[2010 AMC 10A Problems|2010 AMC 10A #9]]}}
 == Problem ==
-A <math>\texti{palindrome}</math>, such as 83438, 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 <math>\text{palindrome}</math>, 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 ==
+===Solution 1===
 <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>.
+===Solution 2===
+For <math>x+32</math> to be a four-digit number, <math>x</math> is in between <math>968</math> and <math>999</math>. The palindromes in this range are <math>969</math>, <math>979</math>, <math>989</math>, and <math>999</math>, so the sum of the digits of <math>x</math> can be <math>24</math>, <math>25</math>, <math>26</math>, or <math>27</math>. Only <math>\boxed{\textbf{(E)}\ 24}</math> is an option, and upon checking, <math>x+32=1001</math> is indeed a palindrome.
+===Solution 3===
+Since we know <math>x+32</math> to be <math>1 a a 1</math> and the only palindrome that works is <math>0 = a</math>, that means <math>x+32 = 1001</math>, and so <math>x = 1001 - 32 = 969</math>. So, <math>9</math> + <math>6</math> + <math>9</math> = <math>\boxed{\textbf{(E)}\ 24}</math>.
+~songmath20
+== Video Solution by OmegaLearn ==
+https://youtu.be/ZhAZ1oPe5Ds?t=1444
+~ pi_is_3.14
+== Video Solution ==
+https://youtu.be/IQj27LEQF4Y
+~Education, the Study of Everything
 == See also ==
 {{AMC12 box|year=2010|num-b=5|num-a=7|ab=A}}
+{{AMC10 box|year=2010|num-b=8|num-a=10|ab=A}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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