Question

The sum of the digits of a three-digit number is $11$. If we subtract $594$ from the number consisting of the same digits written in the reverse order, we shall get a required number. Find that three-digit number, if the sum of all pairwise products of the digits constituting that number is $31$.

Answer 1

Let three digit number be $100x+10y+z$

According to question:

CONDITION 1:

The sum of the digits of a three-digit number is 11

$x+y+z=11\dots \left(1\right)$

CONDITION 2 : the sum of all pairwise products of the digits constituting that number is 31

$xy+yz+zx=31\dots \left(2\right)$

CONDITION 3 : if we subtract $594$ from that number, we shall get a number consisting of the same digits written in the reverse order

$100x+10y+z-594=100z+10y+x$

$99x-99z=594$

$x-z=6$

$x=z+6\dots \left(3\right)$

From $\left(1\right)and\left(3\right)$

$(z+6)+y+z=11$

$\Rightarrow y=5-2z\dots \left(4\right)$

From $\left(2\right),\left(3\right),$ and $\left(4\right)$:

$(z+6)\times (5-2z)+(5-2z)\times \left(z\right)+\left(z\right)\times (z+6)=31$

$5z+30-2z^2-12z+5z-2z^2+z^2+6z=31$

$3z^2-4z+1=0$, which is a quadratic equation

Splitting the middle term:

$3z^2-3z-z+1=0$

$3z(z-1)-1(z-1)=0$

$(3z-1)(z-1)=0$

$z=\frac{1}{3}$ or $z=1$

$z=\frac{1}{3}$ is not possible as digit of any number can't be a fraction

Hence, $z=1$.

From $\left(3\right)$ and $\left(4\right)$:

$y=3$ and $x=7$

Required number $=100\left(7\right)+10\left(3\right)+1=731$

Thus, the required number is $731$.