Difference between revisions of "2016 AMC 8 Problems/Problem 11"
Ethanavery (talk | contribs) m (→Video Solution (CREATIVE THINKING!!!)) |
|||
| Line 4: | Line 4: | ||
<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 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 | ||
==Video Solution (CREATIVE THINKING!!!)== | ==Video Solution (CREATIVE THINKING!!!)== | ||
Revision as of 16:44, 18 June 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 
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
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. 
