Question

The sum of the digits of a three-digit number is $17$ and the sum of the squares of its digits is $109$. If we subtract $495$ from that number, we shall get a number consisting of the same digits written in the reverse order. Find the number.

Answer 1

Let three digit number is $100x+10y+z$

According to question,

CONDITION 1: a sum of the digits of a three-digit number is 17

i.e $x+y+z=\mathrm{17.................}\left(i\right)$

CONDITION 2: the sum of the squares of its digits is 109

i.e $x^2+y^2+z^2=\mathrm{109............}\left(ii\right)$

CONDITION 3: if we subtract 495 from that number, we shall get a number consisting of the same digits written in the reverse order

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

$\Rightarrow 99x-99z=495$

$\Rightarrow x-z=5$

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

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

$(z+5)+y+z=17$

$\Rightarrow y=12-2z...........\left(iv\right)$

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

$(z+5{)}^2+(12-2z{)}^2+z^2=109$

$z^2+25+10z+144+4z^2-48z+z^2=109$

$\Rightarrow 3z^2-19z+30=0$ which is a quadratic equation

on solving we get

$z=\frac{10}{3}andz=3$

$z=\frac{10}{3}$ is not possible as digit of any number can't be a fraction

hence,

$z=3$

now,

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

$y=6andx=8$

thus,

the required number is $863$