Question

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According to question,

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

i.e $x+y+z=17.................(i)$

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

i.e $x_{2}+y_{2}+z_{2}=109............(ii)$

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 $100x+10y+z−495=100z+10y+x$

$⇒99x−99z=495$

$⇒x−z=5$

$⇒x=z+5..............(iii)$

From $(i)and(iii)$

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

$⇒y=12−2z...........(iv)$

From $(i),(iii),and(iv)$

$(z+5)_{2}+(12−2z)_{2}+z_{2}=109$

$z_{2}+25+10z+144+4z_{2}−48z+z_{2}=109$

$⇒3z_{2}−19z+30=0$ which is a quadratic equation

on solving we get

$z=310 andz=3$

$z=310 $ is not possible as digit of any number can't be a fraction

hence,

$z=3$

now,

from $(iii)and(iv)$

$y=6andx=8$

thus,

the required number is $863$

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