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 friend

here is your answer

assuming the number = 10a + b

so,
a × b = 18.


and
10a + b + 63 = 10b + a
9a - 9b = -63
9 ( a - b ) = -9 × 7
a - b = -7


so,
b = 18/a

$a - \frac{18}{a} = - 7 \\ {a}^{2} - 18 = - 7a \\ {a}^{2} + 7a - 18 = 0 \\ {a}^{2} + 9a - 2a - 18 = 0 \\ a(a + 9) - 2(a + 9) = 0 \\ (a + 9)(a - 2) \\ \\ a = - 9(invalid) \: \\ \: \: \: or \: a =2\\ \:$



so, the actual number = 29


glad to help you
hope it helps
thank you.

Answer 2

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

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