Difference between revisions of "2016 AMC 8 Problems/Problem 11"
Pinotation (talk | contribs) |
|||
| (13 intermediate revisions by 10 users not shown) | |||
| Line 5: | Line 5: | ||
<math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math> | <math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math> | ||
| − | + | ==Solution 1== | |
| − | |||
| − | We can | + | We can see that the original number can be written as <math>10a+b</math>, where <math>a</math> represents the tens digit and <math>b</math> represents the units digit. When this number is added to the number obtained by reversing its digits, which is <math>10b+a</math>, the sum would be <math>11a+11b</math>. From this, we can construct the equation <math>11a+11b=132</math>, which simplifies to <math>a+b=12</math>. Since there are 7 pairs of such digits <math>a</math> and <math>b</math>, <math>(3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)</math>, the answer would be <math>\boxed{\textbf{(B) } 7}.</math> |
| − | < | + | |
| − | + | ~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> | ||
| + | |||
| + | ==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 \( 13-6= \) <math>\boxed{\textbf{(B) } 7}.</math>. | ||
| + | |||
| + | ~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 <math>1 + 11, 2 + 10</math> etc. meaning there are <math>\boxed{\textbf{(B) } 7}.</math> solutions. | ||
==Video Solution (CREATIVE THINKING!!!)== | ==Video Solution (CREATIVE THINKING!!!)== | ||
| Line 18: | Line 34: | ||
==Video Solution== | ==Video Solution== | ||
| − | https://youtu.be/ | + | https://youtu.be/lbfbJea43ldk |
~savannahsolver | ~savannahsolver | ||
| Line 26: | Line 42: | ||
{{AMC8 box|year=2016|num-b=10|num-a=12}} | {{AMC8 box|year=2016|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 12:50, 10 October 2025
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.
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
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. 
