Question

A two digit number is such that the product of its digits is 18. When 63 us subtracted from the number, the digits interchange their places. Find the number.

Answer 1

Let the number be 10x+y
A/Q,
xy=18
x=18/y .........(ⅰ)
10x+y-63=10y+x
9x-9y=63
x-y=7
From(ⅰ),we get
18/y-y=7
18-y²/y=7
18-y²=7y
y²+7y-18=0
y²+9y-2y-18=0
y(y+9)-2(y+9)=0
(y-2)(y+9)=0
y=2and-9
If y=2 then x=18/2=9
Number=10×9+2=92
If y=-9 then x=18/-9=-2
Number=10×-2-9=-29

Answer 2

${\green {\boxed {\mathtt {☆Solution}}}}$

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