Question

A two digit no. Is such that the product of its digits is 18. When 63 is subtracted from the no. The digits interchange their places . Find the number

Answer 1

Hey

Here is your answer,

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, which is not possible.
So, x = 2, y = 7

So, the number = 10*9 + 2
= 90 + 2
= 92

Hope it helps you!

Answer 2

$\rm \: let \: the \: tens \: and \: unit \: digit \: of \: the \: required \: number \: be \: x \: and \: y \: respectively \: then \\ \rm \: xy = 18 \implies \: y = \frac{18}{x} \\ \rm \: \purple {\: and \: (10x + y) - 63 = 10y + x }\\ \rm\implies \: 9x - 9y = 63 \implies \: x - y = 7 \: \: \: \: .....(1) \\ \rm \orange{ \: putting \: y = \frac{18}{x } \: into \: (1) }\\ \rm \: x - \frac{18}{x} = 7 \\ \rm \: x {}^{2} - 18 - 7x \implies \: x {}^{2} - 7x - 18 \\ \rm \implies \: x {}^{2} - 9x + 2x - 18 = 0 \implies \: x(x - 9) + 2(x - 9) = 0 \\ \rm \implies(x - 9)(x + 2) = 0 \\ \rm \: x = 9 \: or \: x = - 2 \: \: \: \: ( but \: a \: digit \: cannot \: be \: negative) \\ \rm \red {\: \boxed{\therefore \: x = 9}} \\ \rm \: putting \: x = 9 \: in \: (1) we \: get \: y = 2 \\ \rm \: thus \: the \: tens \: digit \: is \: 9 \: and \: the \: unit \: digit \: is \: 2 \\ \rm hence \: the \: required \: number \: is \: 92$