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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 8"
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== Solution 2 ==
 
== Solution 2 ==
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Denote this number as <math>\overline{ab}</math>.
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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>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>. Yay!!!
  
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Hence, we have <math>\overline{ab} = a + b^2</math>.
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~Andlind
  
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This can be written as <math>10 a + b = a + b^2</math>.
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==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/u19JD85a9d0
  
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Hence, <math>b \left( b - 1 \right) = 9 a</math>.
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~Education, the Study of Everything
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This implies <math>9 | b \left( b - 1 \right)</math>.
 
  
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Hence, either <math>9 | b</math> or <math>9 | b - 1</math>.
 
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Because <math>b \in \left\{ 0 , 1 , \cdots , 9 \right\}</math>, <math>b = 9</math> or 1.
 
  
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For <math>b = 9</math>, we get <math>a = 8</math>. This is a solution.
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==Video Solution by TheBeautyofMath==
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https://youtu.be/ycRZHCOKTVk?t=391
  
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For <math>b = 1</math>, we get <math>a = 0</math>. However, recall that <math>a \neq 0</math>.
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~IceMatrix
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Hence, this is not a solution.
 
  
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Therefore, the answer is <math>\boxed{\textbf{(B) }1}</math>.
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==Video Solution by WhyMath==
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https://youtu.be/knVmshj9SDs ~savannahsolver
  
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~Steven Chen (www.professorchenedu.com)
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==Video Solution by HS Competition Academy==
 +
https://youtu.be/3Ji2_ZYIsPM
  
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== Solution 3 ==
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~Charles 3829
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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>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{1}</math> 2-digit cuddly number, and it is <math>89</math>.
 
  
−
~Andlind
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==Video Solution==
 +
https://youtu.be/6XwPk7h8CU8
 +
 
 +
~Lucas
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=7|num-a=9}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:35, 8 September 2024

Problem

A two-digit positive integer is said to be $\emph{cuddly}$ 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?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution 1

Note that the number $\underline{xy} = 10x + y.$ By the problem statement, \[10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).\] From this we see that $y(y-1)$ must be divisible by $9.$ This only happens when $y=9.$ Then, $x=8.$ Thus, there is only $\boxed{\textbf{(B) }1}$ cuddly number, which is $89.$

~NH14

Solution 2

If the tens digit is $a$ and the ones digit is $b$ then the number is $10a+b$ so we have the equation $10a + b = a + b^2$. We can guess and check after narrowing the possible cuddly numbers down to $13,14,24,25,35,36,46,47,57,68,78,89,$ and $99$. (We can narrow it down to these by just thinking about how $a$'s value affects $b$'s value and then check all the possiblities.) Checking all of these we get that there is only $\boxed{\textbf{(B) }1}$ 2-digit cuddly number, and it is $89$. Yay!!!

~Andlind

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/u19JD85a9d0

~Education, the Study of Everything


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

https://youtu.be/3Ji2_ZYIsPM

~Charles 3829

Video Solution

https://youtu.be/6XwPk7h8CU8

~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

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