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?

Correct Answer: (D) 52

Working backwards: The smallest two-digit number is 10, so D receives 10 coins. Since C gave half of C's coins plus 2 to D, $\frac{C}{2}+2=10⟹\frac{C}{2}=8⟹C=16$$\frac{C}{2}$ + 2 = 10 $\implies$ $\frac{C}{2}$ = 8 $\implies$ C = 16. Similarly, B gave half of B's coins plus 2 to C, so $\frac{B}{2}+2=16⟹\frac{B}{2}=14⟹B=28$$\frac{B}{2}$ + 2 = 16 $\implies$ $\frac{B}{2}$ = 14 $\implies$ B = 28. Finally, A gave half of A's coins plus 2 to B, so $\frac{A}{2}+2=28⟹\frac{A}{2}=26⟹A=52$$\frac{A}{2}$ + 2 = 28 $\implies$ $\frac{A}{2}$ = 26 $\implies$ A = 52.