The given expression is:

$\frac{(1 \times 2 \times 3 \times 4 \times \cdots \times 208) + 3}{30}$

First, notice that $1 \times 2 \times 3 \times \cdots \times 208$ represents the factorial of 208, denoted as $208!$. So, the expression becomes:

$\frac{208! + 3}{30}$

Now, we are interested in the remainder when $208! + 3$ is divided by 30.

Step 1: Analyze $208!$ modulo 30

We know that $208!$ is the product of all integers from 1 to 208. Since $208!$ includes multiples of 2, 3, and 5 (which are factors of 30), it follows that $208!$ is divisible by 30. Therefore:

$208! \mod 30 = 0$

Step 2: Consider $208! + 3$ modulo 30

Since $208! \equiv 0 \mod 30$, we can now calculate:

$(208! + 3) \mod 30 = (0 + 3) \mod 30 = 3$

Thus, the remainder when $(208! + 3)$ is divided by 30 is 3.

Final Answer:

The remainder is 3.

Would you like any additional details or clarification?

Here are 5 related questions to consider:

1. What is the definition of factorial and how is it used in combinatorics?
2. How do you calculate large factorials efficiently?
3. How do modular arithmetic and remainders work in division problems?
4. What are the properties of divisibility rules for numbers like 30?
5. What role do prime factors play in simplifying large factorials?

Tip: For large numbers, breaking the problem into smaller parts, such as focusing on the factors of the divisor, simplifies the process.