What is the remainder when 1!+2!+3!…+100! is divided by 16

To find the remainder when the sum \(1! + 2! + 3! + \dots + 100!\) is divided by 16, we can analyze the factorials modulo 16.

### Step 1: Calculate the factorials modulo 16 for small numbers
- \(1! = 1 \equiv 1 \mod 16\)
- \(2! = 2 \equiv 2 \mod 16\)
- \(3! = 6 \equiv 6 \mod 16\)
- \(4! = 24 \equiv 8 \mod 16\)
- \(5! = 120 \equiv 8 \mod 16\)

### Step 2: Analyze higher factorials modulo 16
For \(n \geq 5\), \(n!\) includes the factor \(16\) because \(n!\) will include a multiple of 16 (since 16 = 2^4 and any factorial with \(n \geq 5\) will have at least four 2's as factors). Therefore:

- \(6! \equiv 0 \mod 16\)
- \(7! \equiv 0 \mod 16\)
- \(\vdots\)
- \(100! \equiv 0 \mod 16\)

### Step 3: Sum the results modulo 16
Only the factorials from \(1!\) to \(4!\) contribute to the sum:

\[
1! + 2! + 3! + 4! = 1 + 2 + 6 + 8 = 17
\]

Now, take this sum modulo 16:

\[
17 \equiv 1 \mod 16
\]

### Conclusion
The remainder when \(1! + 2! + 3! + \dots + 100!\) is divided by 16 is \(\boxed{1}\).

Would you like more details on this, or do you have any questions?

Here are 5 related questions:
1. How would the approach change if we divide by 17 instead of 16?
2. Can you find the remainder when the sum is divided by another number like 9?
3. How does factorial growth affect modular arithmetic properties?
4. What is the smallest value of \(n\) for which \(n!\) is divisible by 100?
5. Can you determine the remainder if the sum only includes factorials from 1! to 50!?

**Tip:** When working with factorials in modular arithmetic, consider how rapidly they grow and what factors they accumulate. This helps in determining when a factorial contributes nothing to the sum under a given modulus.

Discover how to find the remainder when summing factorials from 1! to 100! modulo 16. Detailed step-by-step solution analyzing factorials under modular arithmetic. Suitable for advanced high school students.

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