Find the remainder when `1!+2!+3!+4!++n !` is divided by 15, if `ngeq5.`

Explanation

To solve the problem of finding the remainder when \(1! + 2! + 3! + 4! + \ldots + n!\) is divided by 15 for \(n \geq 5\), we can break it down into steps:

### Step 1: Calculate the factorials from \(1!\) to \(4!\)
We start by calculating the factorials of the first four numbers:
- \(1! = 1\)
- \(2! = 2\)
- \(3! = 6\)
- \(4! = 24\)

### Step 2: Sum the factorials
Now, we sum these factorials:
\[ 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 \]

### Step 3: Consider the factorials from \(5!\) onwards
Next, we analyze the factorials from \(5!\) onwards.
- \(5! = 120\)
- \(6! = 720\)
- \(7! = 5040\)
- And so on.

Notice that \(5!\) and all higher factorials will contain the factors \(3\) and \(5\), which means they will be divisible by \(15\). Therefore, when we divide \(5!\) and higher by \(15\), they will contribute a remainder of \(0\).

### Step 4: Find the remainder of the sum when divided by \(15\)
Since \(5!\) and higher factorials contribute \(0\) to the remainder, we only need to consider the sum \(33\) from \(1! + 2! + 3! + 4!\).

Now, we find the remainder of \(33\) when divided by \(15\):
\[ 33 \div 15 = 2 \quad \text{(quotient)} \]
\[ 33 - (15 \times 2) = 33 - 30 = 3 \quad \text{(remainder)} \]

### Conclusion
Thus, the remainder when \(1! + 2! + 3! + 4! + \ldots + n!\) is divided by \(15\) for \(n \geq 5\) is:
\[ \boxed{3} \]