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 the three digit number if the sum of the pairwise product of the digits constituting the number is 31.

Answer 1

Hint: We are asked to find a three digit number that satisfies the given conditions . given, the sum of digits is 11 , we get $x + y + z = 11$and using third condition we get $x = 6 + z$ and substituting in equation 1 we get the value of y and substituting the values in 2 equation we get value of z , y and x .

Complete step-by-step answer:
We need to find a three digit number that satisfies three conditions
The sum its digits is 11
The sum of the pairwise product of its digits is 31
When we subtract 594 from the number we will get a three digit number with the same digits but in the reversed order.
Let the three digit number be 100x + 10y + z
By condition 1
The sum of its digits is 11
$ \Rightarrow x + y + z = 11$ ………..(1)
By condition 2
The sum of the pairwise product of its digits is 31
$ \Rightarrow xy + yz + zx = 31$ …………(2)
By condition 3
When we subtract 594 from the number we will get a three digit number with the same digits but in the reversed order.
$
   \Rightarrow 100x + 10y + z - 594 = 100z + 10y + x \\
   \Rightarrow 100x + z - 594 = 100z + x \\
   \Rightarrow 100x + z - 594 - 100z - x = 0 \\
   \Rightarrow 99x - 99z - 594 = 0 \\
$
$
   \Rightarrow 99x - 99z = 594 \\
   \Rightarrow 99(x - z) = 594 \\
   \Rightarrow (x - z) = \dfrac{{594}}{{99}} \\
   \Rightarrow x - z = 6 \\
   \Rightarrow x = 6 + z \\
$
Let the above equation be (3)
Substituting this in equation (1)
$
   \Rightarrow 6 + z + y + z = 11 \\
   \Rightarrow 2z + y = 11 - 6 \\
   \Rightarrow 2z + y = 5 \\
$
$ \Rightarrow y = 5 - 2z$ ……….(4)
Substituting (3) and (4) in (2)
$
   \Rightarrow xy + yz + xz = 31 \\
   \Rightarrow (6 + z)(5 - 2z) + (5 - 2z)z + (6 + z)z = 31 \\
   \Rightarrow 30 - 12z + 5z - 2{z^2} + 5z - 2{z^2} + 6z + {z^2} = 31 \\
   \Rightarrow - 3{z^2} + 4z + 30 - 31 = 0 \\
   \Rightarrow 3{z^2} - 4z + 1 = 0 \\
$
By splitting the middle term we get
$
   \Rightarrow 3{z^2} - 3z - z + 1 = 0 \\
   \Rightarrow 3z(z - 1) - 1(z - 1) = 0 \\
   \Rightarrow (3z - 1)(z - 1) = 0 \\
   \Rightarrow z = \dfrac{1}{3}{\text{ or }}z = 1 \\
$
Here we take z = 1 as $z = \dfrac{1}{3}$ is not acceptable as a digit of a number cannot be a fraction
Using z = 1 we get
$
   \Rightarrow x = 6 + z = 6 + 1 = 7 \\
   \Rightarrow y = 5 - 2z = 5 - 2 = 3 \\
$
From this we get the three digit number to be 731.

Note: Many students tend to make a mistake by considering the three digit number to be xyz but it's wrong .
The three digit number is 100x+10y+z
Same way any two digit number will be given by 10x+y