The sum of the squares of the digits constituting a certain positive three-digit number is 74. The hundreds digit of the number is equal to the doubled sum of the digits in the tens and units places. Find the number if it is known that the difference between that number and the number written by the same digits in the reverse order is 495.
Let three digit number is 100x+10y+z
According to question,
CONDITION 1: The sum of the squares of the digits constituting a certain positive three-digit number is 74
74. i.e x2+y2+z2=74.................(i)
CONDITION 2: The hundreds digit of the number is equal to the doubled sum of the digits in the tens and units places
i.e x=2(y+z)............(ii)
CONDITION 3: the difference between that number and the number written by the same digits in the reverse order is 495.
i.e 100x+10y+z−(100z+10y+x)=495
⇒99x−99z=495
⇒x−z=5
⇒x=z+5..............(iii)
From (ii) and (iii)
z+5=2y+2z
⇒y=5−z2...........(iv)
From (i),(iii), and (iv)
(z+5)2+(5−z2)2+z2=74
9z2+30z−171=0
⇒3z2+10−57=0 which is a quadratic equation
on solving we get
z=−193 and z=3
z=−193 is not possible as digit of any number can't be a fraction
hence,
z=3
now,
from (iii) and (iv)
y= 1 and x=8
thus,
the required number is 813