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# 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 $${99(c – a)}$$ = $${297}$$
$${\Rightarrow}$$ $${(c – a)}$$ = $${3}$$

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August 9, 2015
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