Question

The sum of the squares of the digits in a 3 digit number is 98.The sum of the first digit and the last digit is the middle digit. If the digits were reversed,the number is increased by 198.What is the number?

Answer 1

Let a three digit number is xyz
it can be written as 100x + 10y + z

A/C to question,
sum of digits = 98
e.g., x² + y² + z² = 98.......(i)

also sum of 1st and last digit is 2nd digit.
e.g., x + z = y ........(ii)

if the digit were reversed , the number is increased by 198.
e.g., xyz + 198 = zyx
or, 100x + 10y + z + 198 = 100z + 10y + x
or, 99x - 99z + 198 = 0
or, x - z + 2 = 0
or, x = z - 2 .......(iii)

now we have three variables with three equations . let's try to solve it .

from equation (ii) and (iii),
x + z = y
z - 2 + z = y => 2z - 2 = y ......(iv)

now, x² + y² + z² = 98
(z - 2)² + (2z - 2)² + z² = 98 [ from. equation (iii) and (iv) ]

z² - 4z + 4 + 4z² - 8z + 4 + z² = 98
6z² - 12z + 8 = 98
3z² - 6z - 45 = 0
3z² - 15z + 9z - 45 = 0
3z(z - 5) + 9(z - 5) = 0
(3z + 9)(z - 5) = 0
z = 5 and -3 but z ≠ -3 digit should be taken in natural number.
so, z = 5 then, x = 3 and y = 8

now, number is 385