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 ta D.…

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 $x$x.

• A gives half of his coins and 2 more to B. Therefore, the number of coins B receives is:

  $C_B = \frac{x}{2} + 2$CB​=2x​+2

• B then gives half of the coins he received and 2 more to C. The number of coins C receives is:

  $C_C = \frac{C_B}{2} + 2 = \frac{\frac{x}{2} + 2}{2} + 2 = \frac{x}{4} + 1 + 2 = \frac{x}{4} + 3$CC​=2CB​​+2=22x​+2​+2=4x​+1+2=4x​+3

• C subsequently gives half of the coins he received and 2 more to D. The number of coins D receives is:

  $C_D = \frac{C_C}{2} + 2 = \frac{\frac{x}{4} + 3}{2} + 2 = \frac{x}{8} + \frac{3}{2} + 2 = \frac{x}{8} + \frac{7}{2}$CD​=2CC​​+2=24x​+3​+2=8x​+23​+2=8x​+27​

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

Therefore, we set $C_D$CD​ equal to 10:

$\frac{x}{8} + \frac{7}{2} = 10$8x​+27​=10

To solve for $x$x:

$\frac{x}{8} = 10 - \frac{7}{2}$8x​=10−27​

$\frac{x}{8} = \frac{20}{2} - \frac{7}{2}$8x​=220​−27​

$\frac{x}{8} = \frac{13}{2}$8x​=213​

$x = \frac{13}{2} \times 8$x=213​×8

$x = 13 \times 4$x=13×4

$x = 52$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 $x$x into the derived equation $\frac{x}{8} + \frac{7}{2} = 10$8x​+27​=10 does not satisfy the condition. For instance, if $x=76$x=76, then $\frac{76}{8} + \frac{7}{2} = 9.5 + 3.5 = 13 \neq 10$876​+27​=9.5+3.5=13=10. Similarly, for $x=68$x=68, $\frac{68}{8} + \frac{7}{2} = 8.5 + 3.5 = 12 \neq 10$868​+27​=8.5+3.5=12=10, and for $x=60$x=60, $\frac{60}{8} + \frac{7}{2} = 7.5 + 3.5 = 11 \neq 10$860​+27​=7.5+3.5=11=10. Only $x=52$x=52 correctly results in D receiving 10 coins.