The remainder of `(39^(93!))/(40)` is :

Options

A. 0
B. 1
C. 39
D. 13

Answer

To find the remainder of \( \frac{39^{93!}}{40} \), we can use properties of modular arithmetic. Let's break this down step by step.

### Step 1: Understand the problem
We need to find the remainder when \( 39^{93!} \) is divided by \( 40 \).

### Step 2: Use the property of modular arithmetic
We can use the property that states if \( x \equiv -1 \mod n \), then \( x^k \equiv (-1)^k \mod n \). In our case, we can express \( 39 \) in terms of \( 40 \):
\[ 39 \equiv -1 \mod 40 \]
Thus, we can rewrite our expression:
\[ 39^{93!} \equiv (-1)^{93!} \mod 40 \]

### Step 3: Determine the parity of \( 93! \)
Next, we need to determine whether \( 93! \) is odd or even. Since \( 93! \) is the product of all integers from \( 1 \) to \( 93 \), it includes multiple even numbers (like \( 2, 4, 6, \ldots, 92 \)). Therefore, \( 93! \) is even.

### Step 4: Calculate \( (-1)^{93!} \)
Since \( 93! \) is even, we have:
\[ (-1)^{93!} = 1 \]

### Step 5: Conclude the remainder
Thus, we can conclude that:
\[ 39^{93!} \equiv 1 \mod 40 \]
This means that the remainder when \( 39^{93!} \) is divided by \( 40 \) is \( 1 \).

### Final Answer
The remainder of \( \frac{39^{93!}}{40} \) is \( 1 \).

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