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

Revision as of 23:26, 12 December 2021

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?

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

Video Solution

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

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

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 ( $ab$ ) plus the sum of the digits ( $a+b$ ), we can say that $10a+b=ab+a+b$ . We can simplify this to $10a=ab+a$ , 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 ==Video Solution==
 https://youtu.be/7an5wU9Q5hk?t=2226
+https://www.youtube.com/watch?v=RX3BxKW_wTU
 ==Solution==

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