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.
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…(1)
CONDITION 2 : the sum of all pairwise products of the digits constituting that number is 31
xy+yz+zx=31…(2)
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…(3)
From (1) and (3)
(z+6)+y+z=11
⇒y=5−2z…(4)
From (2),(3), and (4):
(z+6)×(5−2z)+(5−2z)×(z)+(z)×(z+6)=31
5z+30−2z2−12z+5z−2z2+z2+6z=31
3z2−4z+1=0, which is a quadratic equation
Splitting the middle term:
3z2−3z−z+1=0
3z(z−1)−1(z−1)=0
(3z−1)(z−1)=0
z=13 or z=1
z=13 is not possible as digit of any number can't be a fraction
Hence, z=1.
From (3) and (4):
y=3 and x=7
Required number =100(7)+10(3)+1=731
Thus, the required number is 731.