A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning?

Assume number of coins with A, B, C and D before giving as A, B, C & D respectively. The number of coins with D = 10 (smallest 2 digit number).

Let’s work backwards now.

D = (C/2) + 2
Or C/2 = 10 – 2 = 8
Or C = 16

Similarly,

C = (B/2) + 2
Or B/2 = 16 – 2 = 14
Or B = 28

And finally,

B = (A/2) + 2
Or A/2 = 28 – 2 = 26
Or A = 52

So, A initially had 52 coins.