Question

I thought of a two digit number. This number is bigger than product of it's digits by 52. What was the number? There is more then one number. PLEASE HELP!! THIS IS DO TODAY!

Answer 1

The number is 73.
7x3+52=73

Answer 2

Let's assume

unit digit is y

tenth digit is x

so, we will get that number as

$=10x+y$

Product of two digits is

$=xy$

now, we have

number is bigger than product of it's digits by 52

so, we get

$10x+y-xy=52$

now, we can solve for y

$y=\frac{52-10x}{1-x}$

now, we know that x and y must be integers

and those numbers must be between 0 and 9

we can use trial and error method

At x=0:

$y=\frac{52-10*0}{1-0}$

$y=52$

it is not possible because y should be between 1 and 9

At x=1:

$y=\frac{52-10*1}{1-1}$

It will make it undefined

so, this is not possible

At x=2:

$y=\frac{52-10*2}{1-2}$

$y=-32$

It is not possible

At x=3:

$y=\frac{52-10*3}{1-3}$

$y=\frac{22}{-2}$

$y=-11$

it is not possible

At x=4:

$y=\frac{52-10*4}{1-4}$

$y=\frac{12}{-3}$

$y=-4$

It is not possible

At x=5:

$y=\frac{52-10*5}{1-5}$

$y=\frac{2}{-4}$

so, it is not possible

At x=6:

$y=\frac{52-10*6}{1-6}$

$y=\frac{-8}{-5}$

It is not possible

At x=7:

$y=\frac{52-10*7}{1-7}$

$y=\frac{-18}{-6}$

$y=3$

so, it is possible

At x=8:

$y=\frac{52-10*8}{1-8}$

$y=\frac{-28}{-7}$

$y=4$

so, it is possible

At x=9:

$y=\frac{52-10*9}{1-9}$

$y=\frac{-38}{-8}$

so, it is not possible

First number is:

x=7 and y=3

$=10*7+3$

$=73$

Second number is:

x=8 and y=4

$=10*8+4$

$=84$

so, two numbers are 73 and 84........Answer