* 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 is unit’s digit, is ten’s digit and is the hundred’s digit.

Then from the given information we get

… (i)

= …(ii)

= …(iii)

=

From (iii) we get =

=

Now from (ii) we get

=

=

So b is a multiple of 3, b can be 0, 3, 6, 9.

For b = 0

not possible as

b = 0 a = 1 c = 4

then

Similarly for b = 3 a = 2 c = 5

& b = 6 a = 3 c = 6 are not possible.

For b = 9 a = 4 c = 7

So the only solution is 497.