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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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==Solution==
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Let the number is <math>\bar{AB}=10A+B</math>, then we have <math>A\times B+A+B=10A+B</math>. <math>A\times(B+1)=10A</math>, <math>B=\box{\texbf{(E)}9}</math>.
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==

Revision as of 21:14, 6 March 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?

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

Solution

Let the number is $\bar{AB}=10A+B$, then we have $A\times B+A+B=10A+B$. $A\times(B+1)=10A$, $B=\box{\texbf{(E)}9}$ (Error compiling LaTeX. Unknown error_msg).

Video Solution by OmegaLearn

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

Video Solution 2

https://www.youtube.com/watch?v=RX3BxKW_wTU ~David

Video Solution 3

https://youtu.be/AR3Ke23N1I8 ~savannahsolver

Video Solution for Problems 21-25

https://www.youtube.com/watch?v=6S0u_fDjSxc

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 $\boxed{\textbf{(E) }9}$

Solution 2

A two digit number is namely $10a+b$, where $a$ and $b$ are digits in which $0 < a \leq 9$ and $0 \leq b \leq 9$. Therefore, we can make an equation with this information. We obtain $10a+b=(a \cdot b) + (a + b)$. This is just $10a+b=ab+a+b.$ Moving $a$ and $b$ to the right side, we get $9a=ab.$ Cancelling out the $a$s, we get $9=b$ which is our desired answer as $b$ is the second digit. Thus the answer is $\boxed{\textbf{(E)}9}$. ~mathboy

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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