Question

a two digit number is such that the product of its digit is 18,wheb 63 is subtracted from the number ,the digits interchange their placed . find the number​

Answer 1

$\huge\underline\bold\red{Answer!!}$

$let \: the \: tens \: digit \: be \: x.$

$then \: the \: units \: digts \: = \frac{18}{x}$

$= > number = 10x + \frac{18}{x}$

and number obtained by interchanging the digits

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 10 \times \frac{18}{x} + x$

$(10x + \frac{18}{x} ) - (10 \times \frac{18}{x} + x) = 63$

$= > 10x + \frac{18}{x} - \frac{180}{x} - x = 0$

$= > 9x - \frac{162}{x} - 63 = 0$

$= > 9x {}^{2} - 63x - 162 = 0$

$= > x {}^{2} - 7x - 18 = 0$

$= > (x - 9)(x + 2) = 0$

$= > x = 9, - 2$

$but \: a \: digit \: can \: never \: be \: ( - ve)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: so,x = 9$

$so \: the \: required \: number \: ,$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 10 \times 9 + \frac{18}{9}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 92$

Hope it helps you ❤️ ✨

Answer 2

Your Answer:

Let the number at unit place be y and at tens place be x

So, the number is 10x+y

ATQ,

xy = 18 ------------->(1)

and

$\tt 10x + y -63 = 10y + x \\\\ \tt \Rightarrow 9x - 9y = 63 \\\\ \tt \Rightarrow 9(x-y) = 63 \\\\ \tt \Rightarrow x-y = \dfrac{63}{9} \\\\ \tt \Rightarrow x-y = 7 \\\\ \tt \Rightarrow x = y +7$

Replacing value of x in Equation 1

$\tt (y+7) y = 18\\\\ \tt \Rightarrow y^2 +7y - 18 = 0 \\\\ \tt \Rightarrow y^2 +9y - 2y -18 = 0 \\\\ \tt \Rightarrow y(y+9) - 2(y+9) = 0 \\\\ \tt \Rightarrow (y-2)(y+9) = 0$

Now equating the factors with zero

y - 2 = 0

=> y = 2

and

y + 9 = 0

=> y = -9

So, y can not be in negative as it is a digit of the number

So, y = 2

and x = 9

And the number is 10x+y = 10(9) + 2 = 90 + 2 = 92

So, the number is 92.