Question

a two digit number is such that the product of the digits is 8 and when 18 is added to the number the digits are reversed

Answer 1

Let the tens and units digit of the required number be 'x' and 'y'
acc.to first condition,
    xy=8
  =>y=8/x
acc.to second condition,
  (10x+y)+18=10y+x
=>10x+y+18=10y+x
=>9x-9y=-18
=>(x-y)=-2
=>x-8/x=-2(from y=8/x)
=>x^2-8=-2x
=>x^2+2x-8=0
=>x^2+4x-2x-8=0
=>x(x+4)-2(x+4)=0
=>(x+4)(x-2)=0
=>(x+4)=0,x-2=0
=>x=-4 or x=2
number cannot be negative so we take x=2 then 
y=8/x=8/2=4 the number is 24 

Answer 2

Answer:

The required number is 24.

Step-by-step explanation:

Let the ten's and unit's digit of the required number be x and 8/x respectively.

Then, this number will be (10x + 8/x)

Number obtained on reversing the digits = (80/x + x)

Accordingly,

$(\frac{10x}{1}+\frac{8}{x})+18=\frac{80}{x}+\frac{x}{1}$

$9x-\frac{72}{x}+18=0$

$9x^2+18x-72=0$

$x^2+2x-8=0$

$x(x+4)-2(x+4)=0$

$(x+4)(x-2)=0$

$x=2\;or\;x=-4$

∴ x is 2 since x ≠ -4.

∴,

Ten's digit = 2

Unit's digit = 8/2 = 4

The required number is 24.