Difference between revisions of "2021 Fall AMC 10A Problems/Problem 8"
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== Problem == | == Problem == | ||
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A two-digit positive integer is said to be <math>\emph{cuddly}</math> 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? | A two-digit positive integer is said to be <math>\emph{cuddly}</math> 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? | ||
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Note that the number <math>\underline{xy} = 10x + y.</math> By the problem statement, <cmath>10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).</cmath> From this we see that <math>y(y-1)</math> must be divisible by <math>9.</math> This only happens when <math>y=9.</math> Then, <math>x=8.</math> Thus, there is only <math>\boxed{\textbf{(B) }1}</math> cuddly number, which is <math>89.</math> | Note that the number <math>\underline{xy} = 10x + y.</math> By the problem statement, <cmath>10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).</cmath> From this we see that <math>y(y-1)</math> must be divisible by <math>9.</math> This only happens when <math>y=9.</math> Then, <math>x=8.</math> Thus, there is only <math>\boxed{\textbf{(B) }1}</math> cuddly number, which is <math>89.</math> | ||
| − | ~ | + | ~NH14 |
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== Solution 2 == | == Solution 2 == | ||
If the tens digit is <math>a</math> and the ones digit is <math>b</math> then the number is <math>10+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>. | If the tens digit is <math>a</math> and the ones digit is <math>b</math> then the number is <math>10+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 |
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
Revision as of 16:41, 2 October 2022
Contents
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, ![\[10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).\]](http://latex.artofproblemsolving.com/7/9/9/79916ece27f39c825bfa425031e66cc209c3ee04.png)
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. 
