The remaider of `(2^(59!))/(255)` is :

Options

A. 0
B. 1
C. 55
D. 5

Answer

To find the remainder of \( \frac{2^{59!}}{255} \), we can use properties of modular arithmetic and the Chinese Remainder Theorem (CRT).

### Step 1: Factor 255
First, we factor 255 into its prime factors:
\[ 255 = 3 \times 5 \times 17 \]

### Step 2: Find \( 2^{59!} \mod 3 \)
To find \( 2^{59!} \mod 3 \), we note that:
\[ 2 \equiv -1 \mod 3 \]
Thus,
\[ 2^{59!} \equiv (-1)^{59!} \mod 3 \]
Since \( 59! \) is even (as it contains all integers from 1 to 59, including even numbers), we have:
\[ (-1)^{59!} \equiv 1 \mod 3 \]

### Step 3: Find \( 2^{59!} \mod 5 \)
Next, we find \( 2^{59!} \mod 5 \). By Fermat's Little Theorem:
\[ 2^{4} \equiv 1 \mod 5 \]
We need to find \( 59! \mod 4 \):
- The factorial \( 59! \) contains many multiples of 4 (specifically, every fourth number), so \( 59! \equiv 0 \mod 4 \).
Thus,
\[ 2^{59!} \equiv 2^{0} \equiv 1 \mod 5 \]

### Step 4: Find \( 2^{59!} \mod 17 \)
Now we find \( 2^{59!} \mod 17 \). Again using Fermat's Little Theorem:
\[ 2^{16} \equiv 1 \mod 17 \]
Next, we need \( 59! \mod 16 \):
- The factorial \( 59! \) contains many multiples of 16, so \( 59! \equiv 0 \mod 16 \).
Thus,
\[ 2^{59!} \equiv 2^{0} \equiv 1 \mod 17 \]

### Step 5: Combine results using CRT
Now we have the following system of congruences:
\[ \begin{align*} 2^{59!} & \equiv 1 \mod 3 \\ 2^{59!} & \equiv 1 \mod 5 \\ 2^{59!} & \equiv 1 \mod 17 \\ \end{align*} \]
Since all three congruences yield the same result, we can combine them:
\[ 2^{59!} \equiv 1 \mod 255 \]

### Conclusion
Thus, the remainder when \( 2^{59!} \) is divided by 255 is:
\[ \boxed{1} \]