Question

A two digit no.is such that Product of its digits is 18 when 63 is added to the number the digits interchanges Thier place. Find the number

Answer 1

Hey

Here is your answer,

Let the two digit number be 10x + y

Now, xy = 18 --- (i) and,

(10x + y) - 63 = 10y + x ---- (ii)

form (ii) we have,

10x + y - 63 = 10y + x
=> 9x - 9y = 63
=> x - y = 7 --- (iii)

Now, (x - y)2 = (x + y)2 - 4xy
=> (x + y)2 = (x - y)2 + 4xy
= (7)2 + 4(18)
= 49 + 72
= 121

=> x + y = 11

so, x - y = 7 and
x + y = 11

on adding we get,

2x = 18
=> x = 9
=> y = 2

Hence the two digit number is

10(9) + 2 = 92

Hope it helps you!

Answer 2

Let the two digit number is 10x + y

Given, the product of its digits = 18

=> xy = 18 ............1

Again, when 63 is subtracted from the number, the digits interchange their places

=> 10x + y - 63 = 10y + x

=> 10x + y - 10y - x = 63

=> 9x - 9y = 63

=> 9(x - y) = 63

=> x - y = 63/9

=> x - y = 7 .............2

Now, (x + y)2 = (x - y)2 + 4xy

=> (x + y)2 = 72 + 4 * 18

=> (x + y)2 = 49 + 72

=> (x + y)2 = 121

=> x + y = √121

=> x + y = ±11

Case 1. when x - y = 7 and x + y = 11

After solving it, we get

x = 9, y = 2

Case 2. when x - y = 7 and x + y = -11

After solving it, we get

x = -2, y = -9, wehich is not possible.

So, x = 2, y = 7

So, the number = 10*9 + 2

                         = 90 + 2

                         = 92