Difference between revisions of "2016 AMC 8 Problems/Problem 11"
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===Solution 2 -SweetMango77=== | ===Solution 2 -SweetMango77=== | ||
Since the numbers are “mirror images,” their average has to be <math>\frac{132}{2}=66</math>. The highest possible value for the tens digit is <math>9</math> because it is a two-digit number. <math>9-6=3</math> and <math>6-3=3</math>, so our lowest tens digit is <math>3</math>. The numbers between <math>9</math> and <math>3</math> inclusive is <math>9-3+1=\boxed{\text{(B)}\;7}</math> total possibilities. | Since the numbers are “mirror images,” their average has to be <math>\frac{132}{2}=66</math>. The highest possible value for the tens digit is <math>9</math> because it is a two-digit number. <math>9-6=3</math> and <math>6-3=3</math>, so our lowest tens digit is <math>3</math>. The numbers between <math>9</math> and <math>3</math> inclusive is <math>9-3+1=\boxed{\text{(B)}\;7}</math> total possibilities. | ||
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+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/G_0KQJhZKGY | ||
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+ | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== |
Revision as of 13:08, 4 April 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 (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.