2014 AMC 8 Problems/Problem 22

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

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2014 AMC 8 Problems/Problem 22

Contents

Problem

Solution

Solution 2

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