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

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==Solution 2==
 
==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>.
+
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>.
 
~mathboy282
 
~mathboy282
  

Revision as of 23:26, 21 October 2020

Problem

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

We can think of the number as $10a+b$, where a and b are digits. Since the number is equal to the product of the digits ($a\cdot b$) plus the sum of the digits ($a+b$), we can say that $10a+b=a\cdot b+a+b$. We can simplify this to $10a=a\cdot b+a$, and factor 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 $\boxed{\textbf{(E)}9}$. ~mathboy282

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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