Difference between revisions of "2016 AMC 8 Problems/Problem 11"
m (→Solution 1) |
|||
Line 10: | Line 10: | ||
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. |
===Solution 2 -SweetMango77=== | ===Solution 2 -SweetMango77=== |
Revision as of 05:32, 17 January 2023
Contents
[hide]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: 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.
Solution 2 -SweetMango77
Since the numbers are “mirror images,” their average has to be . The highest possible value for the tens digit is because it is a two-digit number. , and , so our lowest tens digit is . The numbers between and inclusive is total possibilities.
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.