Question

A two digit number is such that the product of their digits is 12. When 36 is added to this number the digits interchange their places. Determine the number

Answer 1

Let the first digit be x and second digit be y.

Two digits number are in the form of 10x + y

where x is the first digit and y is the second

Given that product of digit is 12

=> xy = 12

Also, given if 36 is added then the digits interchange their places

=> when 36 is added the number is reversed


So the equation is,

10x + y + 36 = 10y + x

(Since, Reverse of 10x + y = 10y + x)

=> 10x + y - 10y - x = -36

=> 9x - 9y = - 36

=> - (9x - 9y) = - (-36)

=> 9y - 9x = 36

=> 9(y - x) = 36

=> y - x = 36/9

=> y - x = 4...................( 1 )

Now squaring both sides,

(y - x)² = (4)²

=> y² + x² - 2xy = 16

Given that

xy = 12

=> 2xy = 24

So,

y² + x² - 2xy = 16

=> y² + x² - 24 = 16

=> y² + x² = 16 + 24

=> y² + x² = 40

Now add 2xy to both sides

=> y² + x² + 2xy = 40 + 2xy

=> (y + x)² = 40 + 24

[since a² + b² + 2ab = (a + b)²]

=> (y + x)² = 64

=> y + x = √64

=> y + x = 8............. ( 2 )

Now add ( 1 ) and ( 2 )

=> y - x + y + x = 4 + 8

=> 2y = 12

=> y = 12/2

=> y = 6

Now put the value of y in any equation to get the value of x. Here we put the value of y in ( 2 )

=> y + x = 8

=> 6 + x = 8

=> x = 8 - 6

=> x = 2


So the number is 10x + y

=> 10(2) + 6

=> 20 + 6

=> 26


So the number is 26

Your answer :- 26

Hope it helps dear friend ☺️

Answer 2

Let,
The tens digit be x
And,
The ones digit be y
Now,
xy=12=>y=12/x ......(i)
Again,
10x+y+36=10y+x ........(ii)
Putting value of y in above equation:-
10x+12/x+36=10*12/x+x
(10x²+12+36x)/x=(120+x²)/x
10x²+36x+12=x²+120
10x²-x²+36x+12-120=0
9x²+36x-108=0
9(x²+4x-12)=0
x²+4x-12=0
x²+6x-2x-12=0
x(x+6)-2(x+6)=0
(x-2)(x+6)=0
Either,
x-2=0
x=2

or

x+6=0
x=-6

Putting values of x in equation (i):-
2y=12
y=6

or

-6y=12
y=-2

Ignoring negative values of x and y:-
Required number=10x+y=10*2+6=26

Hope this helps you!