Difference between revisions of "2016 AMC 8 Problems/Problem 11"
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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: |
Revision as of 10:20, 16 January 2024
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 - kindlymath55532
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.
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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 |
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