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?

The correct option is 52.

Let us understand the problem step-by-step to find the number of coins A had in the beginning. We can solve this problem by working backwards from the final amount of coins D has.

Step 1: Determine the number of coins D has.
The problem states that the number of coins D has is the smallest two-digit number. The smallest two-digit number is 10. Therefore, D has 10 coins.
$D=10$

Step 2: Find the number of coins C had before giving to D.
C gives half of his coins and 2 more to D. Let $C$ be the number of coins C had originally. The number of coins C gives to D is:
$\frac{C}{2}+2$
This expression must equal the number of coins D has, which is 10:
$\frac{C}{2}+2=10$
Subtract 2 from both sides:
$\frac{C}{2}=8$
Multiply by 2:
$C=16$
So, C had 16 coins before giving any to D.

Step 3: Find the number of coins B had before giving to C.
B gives half of his coins and 2 more to C. Let $B$ be the number of coins B had originally. The number of coins B gives to C is:
$\frac{B}{2}+2$
Since C receives these coins and C has 16 coins in total, we can set up the equation:
$\frac{B}{2}+2=16$
Subtract 2 from both sides:
$\frac{B}{2}=14$
Multiply by 2:
$B=28$
So, B had 28 coins before giving any to C.

Step 4: Find the number of coins A had in the beginning.
A gives half of his coins and 2 more to B. Let $A$ be the number of coins A had in the beginning. The number of coins A gives to B is:
$\frac{A}{2}+2$
Since B receives these coins and B has 28 coins in total, we can set up the equation:
$\frac{A}{2}+2=28$
Subtract 2 from both sides:
$\frac{A}{2}=26$
Multiply by 2:
$A=52$

Therefore, A had 52 coins in the beginning, which matches the correct option.