Understanding the Problem: Sum of Factorials and Remainder

The question asks for the remainder when the sum S, defined as the sum of factorials from 0! up to 100!, is divided by 8. The sum is given by:

\(S = 0! + 1! + 2! + 3! + 4! + \dots + 100!\)

To find the remainder when S is divided by 8, we can find the remainder of each term when divided by 8 and then sum these remainders. The final remainder will be the remainder of this sum of remainders when divided by 8.

Calculating Individual Factorial Remainders

Let's calculate the first few factorial values and their remainders when divided by 8:

• \(0! = 1\). Remainder when 1 is divided by 8 is 1.
• \(1! = 1\). Remainder when 1 is divided by 8 is 1.
• \(2! = 2\). Remainder when 2 is divided by 8 is 2.
• \(3! = 3 \times 2 \times 1 = 6\). Remainder when 6 is divided by 8 is 6.
• \(4! = 4 \times 3 \times 2 \times 1 = 24\). \(24 = 3 \times 8 + 0\). Remainder when 24 is divided by 8 is 0.
• \(5! = 5 \times 4! = 5 \times 24 = 120\). \(120 = 15 \times 8 + 0\). Remainder when 120 is divided by 8 is 0.

Identifying the Pattern for Higher Factorials

Notice that \(4!\) is a multiple of 8. Any factorial \(n!\) for \(n \ge 4\) can be written as:

\(n! = n \times (n-1) \times \dots \times 4 \times 3 \times 2 \times 1\)

Since \(n!\) contains the factor \(4!\) for \(n \ge 4\), and \(4!\) is divisible by 8, it means that \(n!\) is also divisible by 8 for all \(n \ge 4\). Therefore, the remainder of \(n!\) when divided by 8 is 0 for all \(n \ge 4\).

Calculating the Sum of Remainders

The sum S can be written as:

\(S = (0! + 1! + 2! + 3!) + (4! + 5! + \dots + 100!)\)

When finding the remainder of S divided by 8, we only need to consider the terms whose remainders are non-zero. From our calculation, only the first four terms (0!, 1!, 2!, and 3!) have non-zero remainders when divided by 8. All terms from 4! up to 100! have a remainder of 0 when divided by 8.

So, the remainder of S when divided by 8 is the same as the remainder of \((0! + 1! + 2! + 3!)\) when divided by 8.

Let's calculate the sum of these terms:

\(0! + 1! + 2! + 3! = 1 + 1 + 2 + 6 = 10\)

Now, we need to find the remainder when 10 is divided by 8.

\(10 \div 8\)

\(10 = 1 \times 8 + 2\)

The remainder is 2.

Thus, the remainder when the sum S is divided by 8 is 2.

Summary of Remainders

Sum of relevant remainders = \(1 + 1 + 2 + 6 = 10\). Remainder of \(10 \div 8\) is 2.

Final Answer Conclusion

The remainder when the sum \(S = 0! + 1! + 2! + 3! + \dots + 100!\) is divided by 8 is 2.

Revision Table: Sum of Factorials Remainder

Additional Information: Factorials and Modulo

Understanding how factorials behave under modular arithmetic is key to solving problems like this. Factorials grow very rapidly. When considering a modulus \(m\), the factorial \(n!\) will eventually contain \(m\) as a factor (if \(m\) is prime or a product of smaller primes present in \(n!\)) or contain factors that multiply to \(m\).

For example, with modulus 8 (which is \(2^3\)):

• 1! has one factor of 2.
• 2! has two factors of 2 (\(2 \times 1\)).
• 3! has one factor of 2 (\(3 \times 2 \times 1\)).
• 4! has three factors of 2 (\(4 = 2^2\), so \(4! = 2^2 \times 3 \times 2 \times 1 = 2^3 \times 3\)). Since it has at least three factors of 2, 4! is divisible by 8.
• For \(n > 4\), \(n!\) includes the factor \(4!\), so it will also be divisible by 8.

This principle applies to other moduli as well. For any composite modulus \(m\), \(n!\) will be divisible by \(m\) for all sufficiently large \(n\).