Question

A two digit number is such that the product of its digits is 18. when 63 is subtracted from the number, the digits interchamge their place. find the number.​

Answer 1

Answer:

$\bf \underline{The\ original\ number\ is\ 92.}$

Step-by-step explanation:

Let the digit at tens and units place be x and y respectively.

Amd the number be (10x+y)

Given

Product of digit is 18.

So,

ATE, xy=18

or y=18/x -(1)

Also, when 63 is subtracted from the number, the digits interchange their place.

So, the number formed by interchanging the digits is (10y+x)

ATE,

(10x+y)-63=(10y+x)

10x-x+y-10y=63

9x-9y=63

9(x-y)=63

x-y=7 -(2)

Using y=18/x in eq. (2)

$x-\frac{18}{x}=7\\ \\ \implies \rm x^2-18=7\\ \\ \implies \rm \frac{x^2-18}{x}=7\\ \\ \implies \rm x^2-7x-18=0\\ \\ \implies \rm x^2-9x+2x-18=0\\ \\ \implies \rm x(x-9) 2(x-9)=0\\ \\ \implies \rm (x+2) (x-9)=0\\ \\ \implies \bf x=-2\ or\ \bf 9 \\ \rm But\ digit\ can't\ be\ negative.\\ \therefore \bf x=9$

Putting x=9 in eq. (1)

9y=18

y=2

So, the number is (10×9+2)=90+2=92

Answer 2

Hey there

refer to attachment