@Rohit-Rathore

Q3) Let say we had x coins in the beginning

After 1st loot : x - 1 - (x-1)/3 = 2/3(x-1) = (2x-2)/3

After 2nd loot : (2x - 2)/3 - 1 - [(2x - 2)/3 - 1]/3 = 2/3[(2x-2)/3 - 1] = (4x - 10)/9

After 3rd loot : (4x - 10)/9 - 1 - [((4x - 10)/9 - 1]/3 = 2/3 [(4x - 10)/9 - 1] = (8x - 38)/27

Now the final value (after throwing away the fake coin) = (8x - 38)/27 - 1 = (8x - 65)/27 is a multiple of 3.

So we can write, (8x - 65)/27 = 3k

x = (81k + 65)/8

x is an integer, 81K + 65 should be a multiple of 8, means (81K + 65) Mod 8 = 0

We know 81 Mod 8 = 1 and 65 mod 8 = 1. So to get remainder 0, we need to adjust K in a way that 81K mod 8 = 7.

Least value would be K = 7

For K = 7, x = (81 * 7 + 65)/8 = 79, which should (could) be our answer.

Let me know in case of any Logical/Calculation errors.

DILR - Booster