# Sum of squares of digits (Tomato subjective 74)

problem: The sum of squares of the digits of a three-digit  positive number is 146, while the sum of the two digits in the unit’s and the ten’s place is 4 times the digit in the hundred’s place. Further, when the number is written in the reverse order, it is increased by 297. Find the number.

solution: Let $${c}$$ is unit’s digit, $${b}$$ is ten’s digit and $${a}$$ is the hundred’s digit.

Then from the given information we get

$${\displaystyle{a^2 + b^2 + c^2 = 146}}$$     … (i)
$${b + c}$$ = $${4a}$$ …(ii)
$${100a + 10b + c + 297}$$ = $${100c + 10 + a}$$ …(iii)
$${b + c}$$ = $${4a}$$
From (iii) we get $${99(c – a)}$$ = $${297}$$
$${\Rightarrow}$$ $${(c – a)}$$ = $${3}$$
Now from (ii) we get

$${\Rightarrow}$$ $${b + 3}$$ = $${3a}$$
$${\Rightarrow}$$ $${b}$$ = $${3 (a – 1)}$$
So b is a multiple of 3, b can be 0, 3, 6, 9.
For b = 0
not possible as
b = 0 $${\Rightarrow}$$ a = 1 $${\Rightarrow}$$ c = 4
then $${\displaystyle{a^2 + b^2 + c^2 {\ne} 146}}$$
Similarly for b = 3 $${\Rightarrow}$$ a = 2 $${\Rightarrow}$$ c = 5
& b = 6 $${\Rightarrow}$$ a = 3 $${\Rightarrow}$$ c = 6 are not possible.
For b = 9 $${\Rightarrow}$$ a = 4 $${\Rightarrow}$$ c = 7
$${\displaystyle{a^2 + b^2 + c^2 = 146}}$$
So the only solution is 497.