There are four persons A, B, C, D; and A has some coins. A gave half of the coins to B and 4 more besides. B gave half of the coins to C and 4 more besides. C gave half of the coins to D and 4 more besides. Both B and D end up with same number of coins. How many coins did A have originally

Explanation

The question involves a series of transactions where each person gives half of their coins, plus 4 more, to the next person. For option 1: If A had initially 96 coins, then B would have received 48 + 4 = 52. B then gives C half of these coins plus 4 more, which is 26 + 4 = 30, not leaving B and D with the same amount of coins. For option 2: If A originally had 84 coins, B would have gotten 42 + 4 = 46. B passes on 23 + 4 = 27 coins to C, so again, B and D would not have the same number of coins. For option 3: if A had 72 coins, B would receive 36 + 4 = 40. B then gives C 20 + 4 = 24 coins, leaving B with 16 coins. C finally gives D half of those coins plus 4, i.e. 12 + 4 = 16. Hence, B and D both end up with 16 coins, verifying option 3 as the correct solution. For option 4, similarly logic as other options can be used to discount this as a viable