* 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.