The remainder when `(888!)^(9999)` is divided by 77 is :

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To find the remainder when \((888!)^{9999}\) is divided by \(77\), we can use the properties of factorials and modular arithmetic.

### Step 1: Factor the modulus
The first step is to factor \(77\) into its prime factors:
\[ 77 = 7 \times 11 \]

### Step 2: Use the Chinese Remainder Theorem
We will find the remainders of \((888!)^{9999}\) modulo \(7\) and \(11\) separately, and then combine the results using the Chinese Remainder Theorem.

### Step 3: Calculate \(888! \mod 7\)
Since \(888\) is greater than \(7\), \(888!\) contains all integers from \(1\) to \(888\), including multiples of \(7\). Therefore:
\[ 888! \equiv 0 \mod 7 \]
This means that:
\[ (888!)^{9999} \equiv 0^{9999} \equiv 0 \mod 7 \]

### Step 4: Calculate \(888! \mod 11\)
Similarly, since \(888\) is greater than \(11\), \(888!\) contains all integers from \(1\) to \(888\), including multiples of \(11\). Thus:
\[ 888! \equiv 0 \mod 11 \]
This means that:
\[ (888!)^{9999} \equiv 0^{9999} \equiv 0 \mod 11 \]

### Step 5: Combine results using the Chinese Remainder Theorem
Now we have:
\[ (888!)^{9999} \equiv 0 \mod 7 \]
\[ (888!)^{9999} \equiv 0 \mod 11 \]
Since both remainders are \(0\), we can conclude that:
\[ (888!)^{9999} \equiv 0 \mod 77 \]

### Final Answer
Thus, the remainder when \((888!)^{9999}\) is divided by \(77\) is:
\[ \boxed{0} \]