In a two digit number, the sum of the digits is equal to the product of the digits find the number.

Let the digits of the required number be x and y.
Now, the required number is 10x + y.
According to the question,
10x + y = 4(x + y)                
So,
6x − 3y = 0

\[\Rightarrow\]2x − y = 0

\[x = \frac{y}{2}\]                                               .....(1)

Also, 
10x + y = 3xy                                            .....(2)
From (1) and (2), we get

\[10\left( \frac{y}{2} \right) + y = 3\left( \frac{y}{2} \right)y\]
\[ \Rightarrow 5y + y = \frac{3}{2} y^2 \]
\[ \Rightarrow 6y = \frac{3}{2} y^2 \]

\[\Rightarrow y^2 - 4y = 0\]
\[ \Rightarrow y(y - 4) = 0\]
\[ \Rightarrow y = 0, 4\]

So, x = 0 for y = 0 and x = 2 for y = 4.

Hence, the required number is 24. 

Contents

• 1 Problem 22
• 2 Video Solution
• 3 Video Solution for Problems 21-25
• 4 Solution
• 5 Solution 2
• 6 See Also

A

-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?

Video Solution

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

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

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

, where a and b are digits. Since the number is equal to the product of the digits () plus the sum of the digits (), we can say that . We can simplify this to , which factors to . Dividing by , we have that . Therefore, the units digit, , is

Solution 2

A two digit number is namely

, where and are digits in which and . Therefore, we can make an equation with this information. We obtain . This is just Moving and to the right side, we get Cancelling out the s, we get which is our desired answer as is the second digit. Thus the answer is . ~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

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

Which two

What is the sum of the digits of the product?

How many two