To determine the initial number of coins A had, we can set up equations based on the sequential distribution of coins.

Correct Option

Let the initial number of coins A possessed be $x$ .

A gives half of his coins and 2 more to B. Therefore, the number of coins B receives is:
$C_{B} = \frac{x}{2} + 2$
B then gives half of the coins he received and 2 more to C. The number of coins C receives is:
$C_{C} = \frac{C_{B}}{2} + 2 = \frac{\frac{x}{2} + 2}{2} + 2 = \frac{x}{4} + 1 + 2 = \frac{x}{4} + 3$
C subsequently gives half of the coins he received and 2 more to D. The number of coins D receives is:
$C_{D} = \frac{C_{C}}{2} + 2 = \frac{\frac{x}{4} + 3}{2} + 2 = \frac{x}{8} + \frac{3}{2} + 2 = \frac{x}{8} + \frac{7}{2}$

The problem states that the number of coins D has is the smallest two-digit number, which is 10.

Therefore, we set $C_{D}$ equal to 10:

$\frac{x}{8} + \frac{7}{2} = 10$

To solve for $x$ :

$\frac{x}{8} = 10 - \frac{7}{2}$

$\frac{x}{8} = \frac{20}{2} - \frac{7}{2}$

$\frac{x}{8} = \frac{13}{2}$

$x = \frac{13}{2} \times 8$

$x = 13 \times 4$

$x = 52$

Thus, A had 52 coins in the beginning.

Incorrect Options

Options 1 (76), 2 (68), and 3 (60) are incorrect because substituting these values for $x$ into the derived equation $\frac{x}{8} + \frac{7}{2} = 10$ does not satisfy the condition. For instance, if $x = 76$ , then $\frac{76}{8} + \frac{7}{2} = 9.5 + 3.5 = 13 \neq 10$ . Similarly, for $x = 68$ , $\frac{68}{8} + \frac{7}{2} = 8.5 + 3.5 = 12 \neq 10$ , and for $x = 60$ , $\frac{60}{8} + \frac{7}{2} = 7.5 + 3.5 = 11 \neq 10$ . Only $x = 52$ correctly results in D receiving 10 coins.