2016 AMC 8 Problems/Problem 11

Problem

Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$

$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution 1

We can see that the original number can be written as $10a+b$ , where $a$ represents the tens digit and $b$ represents the units digit. When this number is added to the number obtained by reversing its digits, which is $10b+a$ , the sum would be $11a+11b$ . From this, we can construct the equation $11a+11b=132$ , which simplifies to $a+b=12$ . Since there are 7 pairs of such digits $a$ and $b$ , $(3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)$ , the answer would be $\boxed{\textbf{(B) } 7}.$

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Solution 2

We can set the number as ab where a is the tens digit and b is the ones digit. So now the equation will be ab+ba=132. a+b has to have a remainder of 2 when divided by 10 so it will be $a + b \equiv 2 (mod 10)$ . We also know a<10 and b<10. So a+b can either be 2 or 12. a+b cannot be 2 because then there will be only 3 numbers that work and that isn't in the answer choice. So a+b=12 . To check this we can do ab+ba=132 which equals to (a+b)0+(a+b)=132 and since we said a+b=12 we get 120+12=132 which is true. So we have a+b=12 and a<10 and b<10.If a is 9 then b=3 and if a=3 then b=9 calculating how many pairs are in between you get 7. So the answer is $\boxed{\textbf{(B) } 7}.$

Video Solution (CREATIVE THINKING!!!)

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Video Solution

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2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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