Question

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

Answer 1

Answer:

Step-by-step explanation:

Let the digits be X and Y

Now

X.Y=18.......( i )

10X+Y-63=10Y+X

9X-63=9Y

9X-9Y=63

X-Y=7.......( ii )

From ( i ), Y=18/x

So,

X-18/X=7

X^2-18=7X

X^2-7X-18=0

X^2-(9-2)X-18=0

X^2-9X+2X-18=0

X(X-9) + 2(X-9)=0

X+2=0 , X-9=0

So,

X=9

Y=2

Required number is 92

Answer 2

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