one of four #s where n! has n digits.
To find one of the four numbers where n! (n factorial) has n digits, we can use a simple iterative process.
First, let's understand the relationship between n! and the number of digits it has. The number of digits in a positive integer n can be found using the logarithm function. Specifically, the number of digits is given by the floor of the logarithm base 10 of the number, plus 1.
So, we need to find the value of n where the number of digits in n! is equal to n.
Here's the step-by-step process to find one of the four numbers:
1. Start with n = 1.
2. Calculate n! (the factorial of n).
3. Calculate the number of digits in n! using the formula: floor(log10(n!)) + 1.
4. Compare the number of digits with n.
5. If they are equal, then we have found one of the numbers. If not, increase n by 1 and repeat from step 2.
6. Repeat steps 2-5 until we find four numbers where n! has n digits.
Here's a Python code example to find the four numbers:
```python
from math import factorial, log10, floor
count = 0
n = 1
while count < 4:
fact = factorial(n)
digits = floor(log10(fact)) + 1
if digits == n:
print(f"One of the numbers where n! has n digits: {n}! = {fact}")
count += 1
n += 1
```
When you run this code, it will print out the four numbers where n! has n digits.