Question

A two digit number in such that the product of its digits is $8$. When $63$ is subtracted from the number, the digits interchange their places. Find the number.

• A. 18
• B. 72
• C. 27
• D. 81

Answer 1

Answer: (D)

81

Let the digit at unit's place $=x$
$\therefore$ Digit at ten's place $\frac{8}{x}$ and the number is $(\frac{80}{x}+x)$
New number on interchanging the places of digits $=10x+\frac{8}{x}$
$\therefore$ According to given condition
$\frac{80}{x}+x-63=10x+\frac{8}{x}$
$80+x^2-63x=10x^2+8$
${9x}^2+63x-72=0$
$x^2-7x-8=0$
$x^2+8x-x-8=0$
$x(x+8)-1(x+8)=0$
$(x+8)(x-1)=0$
$i.e.x=-8$ and $x=1$
Rejecting $x=-8$ and putting $x=1$ the required no. is $(\frac{80}{1}+1)=81$.