UPSC CDS 2 2021 Math Paper

Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$. The product of the digits is 8 . $\therefore \;\; xy=8 \;\; \ldots \text{ (i) }$ After interchanging the digits, the number becomes $10 x+y$. If 63 is added to the number, the digits interchange their places. Thus, $\;(10 y+x)+63=10 x+y$ $\Rightarrow\;10 x+y-10 y-x=63$ $\Rightarrow\;9 x-9 y=63$ $\Rightarrow\;x-y=7$ . . . (ii) From Eq. (ii), put $y=x-7$ into Eq. (i), we get $\;x(x-7) \;\; =8$ $\Rightarrow\;x^{2}-7 x-8 \;\; =0$ $\Rightarrow\;x^{2}-8 x+x-8 \;\; =0$ $\Rightarrow\;x(x-8)+1(x-8) \;\; =0$ $\Rightarrow\;\;(x-8)(x+1)=0\;\Rightarrow\;x=8 \;\text{ or }\;-1$ From Eq. (ii), when $x=8$ $\Rightarrow \;\; y=8-7=1$ When $x=-1$ $\Rightarrow \;\; y=-1-7=-8$ We get $(x, y)=(8,1)$ and $(x, y)=(-1,-8)$ Since, the digits of the number can't be negative. So, we must remove second pair. Therefore, the number is $10 \times 1+8=18$ Hence, sum of the digits in the number is $1+8=9$.