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Difference between revisions of "2014 AMC 8 Problems/Problem 22"

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<math>\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }9</math>
 
<math>\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }9</math>
  
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==Video Solution==
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==Video Solution by OmegaLearn==
 
https://youtu.be/7an5wU9Q5hk?t=2226
 
https://youtu.be/7an5wU9Q5hk?t=2226
  
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https://www.youtube.com/watch?v=RX3BxKW_wTU
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==Video Solution 3==
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https://youtu.be/AR3Ke23N1I8 ~savannahsolver
 
https://youtu.be/AR3Ke23N1I8 ~savannahsolver
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==Video Solution for Problems 21-25==
 
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https://www.youtube.com/watch?v=6S0u_fDjSxc
 
  
 
==Solution==
 
==Solution==
 
We can think of the number as <math>10a+b</math>, where a is the tens digit and b is the unit digit. Since the number is equal to the product of the digits (<math>ab</math>) plus the sum of the digits (<math>a+b</math>), we can say that <math>10a+b=ab+a+b</math>. We can simplify this to <math>10a=ab+a</math>, which factors to <math>(10)a=(b+1)a</math>. Dividing by <math>a</math>, we have that <math>b+1=10</math>. Therefore, the units digit, <math>b</math>, is <math>\boxed{\textbf{(E) }9}</math>
 
We can think of the number as <math>10a+b</math>, where a is the tens digit and b is the unit digit. Since the number is equal to the product of the digits (<math>ab</math>) plus the sum of the digits (<math>a+b</math>), we can say that <math>10a+b=ab+a+b</math>. We can simplify this to <math>10a=ab+a</math>, which factors to <math>(10)a=(b+1)a</math>. Dividing by <math>a</math>, we have that <math>b+1=10</math>. Therefore, the units digit, <math>b</math>, is <math>\boxed{\textbf{(E) }9}</math>
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==Solution 2==
 
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A two digit number is namely <math>10a+b</math>, where <math>a</math> and <math>b</math> are digits in which <math>0 < a \leq 9</math> and <math>0 \leq b \leq 9</math>. Therefore, we can make an equation with this information. We obtain <math>10a+b=(a \cdot b) + (a + b)</math>. This is just <math>10a+b=ab+a+b.</math> Moving <math>a</math> and <math>b</math> to the right side, we get <math>9a=ab.</math> Cancelling out the <math>a</math>s, we get <math>9=b</math> which is our desired answer as <math>b</math> is the second digit. Thus the answer is <math>\boxed{\textbf{(E)}9}</math>.
 
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~mathboy
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=21|num-a=23}}
 
{{AMC8 box|year=2014|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:51, 23 July 2024

Problem 22

A $2$-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?

$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }9$

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=2226

Video Solution 3

https://youtu.be/AR3Ke23N1I8 ~savannahsolver

Solution

We can think of the number as $10a+b$, where a is the tens digit and b is the unit digit. Since the number is equal to the product of the digits ($ab$) plus the sum of the digits ($a+b$), we can say that $10a+b=ab+a+b$. We can simplify this to $10a=ab+a$, which factors to $(10)a=(b+1)a$. Dividing by $a$, we have that $b+1=10$. Therefore, the units digit, $b$, is $\boxed{\textbf{(E) }9}$

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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