Difference between revisions of "2016 AMC 8 Problems/Problem 11"

Latest revision as of 13:30, 28 July 2025

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 $\boxed{\textbf{(B) } 7}.$

~Aqf243

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 $a+b\equiv 2\quad (\bmod 10)$. 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!!!)

https://youtu.be/G_0KQJhZKGY

~Education, the Study of Everything

Video Solution

https://youtu.be/lbfbJea43ldk

~savannahsolver

@@ Line 10: / Line 10: @@
 ~Aqf243
+==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 \(a+b\equiv 2\quad (\bmod 10)\). 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 <math>\boxed{\textbf{(B) } 7}.</math>
 ==Video Solution (CREATIVE THINKING!!!)==

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