Difference between revisions of "2021 Fall AMC 10A Problems/Problem 8"
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− | If the tens digit is <math>a</math> and the ones digit is <math>b</math> then the number is <math> | + | If the tens digit is <math>a</math> and the ones digit is <math>b</math> then the number is <math>10a+b</math> so we have the equation <math>10a + b = a + b^2</math>. We can guess and check after narrowing the possible cuddly numbers down to <math>13,14,24,25,35,36,46,47,57,68,78,89,</math> and <math>99</math>. (We can narrow it down to these by just thinking about how <math>a</math>'s value affects <math>b</math>'s value and then check all the possiblities.) Checking all of these we get that there is only <math>\boxed{\textbf{(B) }1}</math> 2-digit cuddly number, and it is <math>89</math>. |
~Andlind | ~Andlind |
Revision as of 14:35, 13 November 2022
Contents
[hide]Problem
A two-digit positive integer is said to be if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?
Solution 1
Note that the number By the problem statement, From this we see that must be divisible by This only happens when Then, Thus, there is only cuddly number, which is
~NH14
Solution 2
If the tens digit is and the ones digit is then the number is so we have the equation . We can guess and check after narrowing the possible cuddly numbers down to and . (We can narrow it down to these by just thinking about how 's value affects 's value and then check all the possiblities.) Checking all of these we get that there is only 2-digit cuddly number, and it is .
~Andlind
Video Solution by TheBeautyofMath
https://youtu.be/ycRZHCOKTVk?t=391
~IceMatrix
Video Solution by WhyMath
https://youtu.be/knVmshj9SDs ~savannahsolver
Video Solution by HS Competition Academy
~Charles 3829
Video Solution
~Lucas
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.