2016 AMC 8 Problems/Problem 11
Contents
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 
Solution 1
We can see that the original number can be written as 
, where 
represents the tens digit and 
represents the units digit. When this number is added to the number obtained by reversing its digits, which is 
, the sum would be 
. From this, we can construct the equation 
, which simplifies to 
. Since there are 7 pairs of such digits and , 
, the answer would be 
~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
Solution 3
Like similar solutions, we can say that \( 10a+b \) is a two digit number, and therefore we have \( 10a+b + 10b+a = 132 \) which implies \( 11a + 11b = 132 \) which implies \( a + b = 12 \).
We then use stars and bars to find out that there are \( \binom{13}{1} = 13 \) different possible sums.
However, we want to satisfy \( 0 \le a \le 9 \) and \( 0 \le b \le 9 \), therefore we see that the sums \( 0+12 \), \( 1+11 \), \( 2+10 \), and \( 12+0 \), \( 11+1 \), and \( 10+2 \) do not work, and our answer is \( 13-6= \) .
~Pinotation
JasonDaGoat Solution
The two digit number 10A+B + the reversed number 10B+A is equal to 11A+11B = 132. Dividing both sides by 11 gives A + B = 12. Since we can't have 
etc. meaning there are solutions.
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See Also
| 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. 
