A thief somehow managed to steal some golden coins from a bank's cash but while coming out of it at the first door he was caught by the watch man and he successfully dealt him by paying 1 coin plus half of the rest coins. Further he had to pay 2 coins, hten half of the rest to the second watchman. Once again at the rest to the second watchman. Once agian at the third gate (outermost) he gave 3 coins and then half of the rest. After it he was left with only coin. How many coins had he stolen?

Options

A. 32
B. 36
C. 25
D. none of these

Answer

To solve the problem step by step, let's denote the initial number of coins the thief stole as \( x \).

### Step 1: First Watchman
When the thief reaches the first watchman, he pays 1 coin plus half of the remaining coins.

- Coins paid to the first watchman = \( 1 + \frac{x - 1}{2} \)
- Remaining coins after paying the first watchman:
\[ x - \left(1 + \frac{x - 1}{2}\right) = x - 1 - \frac{x - 1}{2} = x - 1 - \frac{x}{2} + \frac{1}{2} = \frac{x}{2} - \frac{1}{2} \]

### Step 2: Second Watchman
Now, the thief reaches the second watchman and pays 2 coins plus half of the remaining coins.

- Remaining coins before the second watchman = \( \frac{x}{2} - \frac{1}{2} \)
- Coins paid to the second watchman = \( 2 + \frac{\left(\frac{x}{2} - \frac{1}{2}\right) - 2}{2} \)
- Remaining coins after paying the second watchman:
\[ \left(\frac{x}{2} - \frac{1}{2}\right) - \left(2 + \frac{\left(\frac{x}{2} - \frac{1}{2} - 2\right)}{2}\right) \]

Calculating the payment:
\[ = 2 + \frac{\frac{x}{2} - \frac{1}{2} - 2}{2} = 2 + \frac{x/2 - 5/2}{2} = 2 + \frac{x - 5}{4} \]
So, the remaining coins:
\[ \frac{x}{2} - \frac{1}{2} - \left(2 + \frac{x - 5}{4}\right) \]
Simplifying:
\[ = \frac{x}{2} - \frac{1}{2} - 2 - \frac{x}{4} + \frac{5}{4} = \frac{2x - 1 - 8 - x + 5}{4} = \frac{x - 4}{4} \]

### Step 3: Third Watchman
Now, the thief reaches the third watchman and pays 3 coins plus half of the remaining coins.

- Remaining coins before the third watchman = \( \frac{x - 4}{4} \)
- Coins paid to the third watchman = \( 3 + \frac{\left(\frac{x - 4}{4} - 3\right)}{2} \)
- Remaining coins after paying the third watchman:
\[ \frac{x - 4}{4} - \left(3 + \frac{\left(\frac{x - 4}{4} - 3\right)}{2}\right) \]

Calculating the payment:
\[ = 3 + \frac{\frac{x - 4}{4} - 3}{2} = 3 + \frac{x - 4 - 12}{8} = 3 + \frac{x - 16}{8} \]
So, the remaining coins:
\[ \frac{x - 4}{4} - \left(3 + \frac{x - 16}{8}\right) \]
Simplifying:
\[ = \frac{x - 4}{4} - 3 - \frac{x - 16}{8} \]
Finding a common denominator (8):
\[ = \frac{2(x - 4)}{8} - \frac{24}{8} - \frac{x - 16}{8} = \frac{2x - 8 - 24 - x + 16}{8} = \frac{x - 16}{8} \]

### Final Step: Remaining Coins
After passing the third watchman, he is left with 1 coin:
\[ \frac{x - 16}{8} = 1 \]
Multiplying both sides by 8:
\[ x - 16 = 8 \]
Adding 16 to both sides:
\[ x = 24 \]

### Conclusion
The thief initially stole **24 coins**.