Difference between revisions of "2016 AMC 8 Problems/Problem 11"
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==Solutions== | ==Solutions== | ||
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===Solution 1=== | ===Solution 1=== | ||
We can write the two digit number in the form of <math>10a+b</math>; reverse of <math>10a+b</math> is <math>10b+a</math>. The sum of those numbers is: | We can write the two digit number in the form of <math>10a+b</math>; reverse of <math>10a+b</math> is <math>10b+a</math>. The sum of those numbers is: | ||
<cmath>(10a+b)+(10b+a)=132</cmath><cmath>11a+11b=132</cmath><cmath>a+b=12</cmath> | <cmath>(10a+b)+(10b+a)=132</cmath><cmath>11a+11b=132</cmath><cmath>a+b=12</cmath> | ||
| − | We can use brute force to find order pairs <math>(a,b)</math> such that <math>a+b=12</math>. Since <math>a</math> and <math>b</math> are both digits, both <math>a</math> and <math>b</math> have to be integers less than <math>10</math>. Thus our ordered pairs are <math>(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)</math> or <math>\boxed{\textbf{(B)} 7}</math> ordered pairs. | + | We can use brute force to find order pairs <math>(a,b)</math> such that <math>a+b=12</math>. Since <math>a</math> and <math>b</math> are both digits, both <math>a</math> and <math>b</math> have to be integers less than <math>10</math>. Thus, our ordered pairs are <math>(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)</math>; or <math>\boxed{\textbf{(B)} 7}</math> ordered pairs. |
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| + | -kindlymath55532 | ||
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| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/G_0KQJhZKGY | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/lbfbJWcVsEE | ||
| − | + | ~savannahsolver | |
| − | |||
==See Also== | ==See Also== | ||
Latest revision as of 16:45, 7 July 2024
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 
Solutions
Solution 1
We can write the two digit number in the form of 
; reverse of is 
. The sum of those numbers is:
![\[(10a+b)+(10b+a)=132\]](http://latex.artofproblemsolving.com/6/f/d/6fd900ef889d9862c522232dbcf8838063afa806.png)
![\[11a+11b=132\]](http://latex.artofproblemsolving.com/8/7/3/8739636c62a0fe31852e4bc3a20648d3ab312286.png)
![\[a+b=12\]](http://latex.artofproblemsolving.com/6/0/5/6059e0bfc25919547d7e4be50406ff6eba53885d.png)
We can use brute force to find order pairs 
such that 
. Since 
and 
are both digits, both and have to be integers less than 
. Thus, our ordered pairs are 
; or 
ordered pairs.
-kindlymath55532
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
