Question

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 $495$.

Answer 1

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 $x^2+y^2+z^2=\mathrm{74.................}\left(i\right)$

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 $x=2(y+z)............\left(ii\right)$

CONDITION 3: the difference between that number and the number written by the same digits in the reverse order is 495.

i.e $100x+10y+z-(100z+10y+x)=495$

$\Rightarrow 99x-99z=495$

$\Rightarrow x-z=5$

$\Rightarrow x=z+\mathrm{5..............}\left(iii\right)$

From $\left(ii\right)and\left(iii\right)$

$z+5=2y+2z$

$\Rightarrow y=\frac{5-z}{2}...........\left(iv\right)$

From $\left(i\right),\left(iii\right),and\left(iv\right)$

$(z+5{)}^2+(\frac{5-z}{2}{)}^2+z^2=74$

$9z^2+30z-171=0$

$\Rightarrow 3z^2+10-57=0$ which is a quadratic equation

on solving we get

$z=\frac{-19}{3}andz=3$

$z=\frac{-19}{3}$ is not possible as digit of any number can't be a fraction

hence,

$z=3$

now,

from $\left(iii\right)and\left(iv\right)$

$y=1andx=8$

thus,

the required number is $813$