he-024

1.000

1/1 tests· basic

Proposed solution

Looking at this problem, I need to find the largest divisor of `n` that is smaller than `n` itself.

Let me think through the approach:
1. For any number n > 1, I want to find the largest proper divisor
2. The largest proper divisor would be n divided by its smallest prime factor
3. For example, with n=15: factors are 1,3,5,15. The largest proper divisor is 5 (which is 15/3)
4. For a prime number like 7: the only proper divisor is 1

I can optimize this by checking divisors from sqrt(n) downwards to find the largest one efficiently.

Actually, let me think more carefully:
- I want the largest divisor d where d < n and n % d == 0
- This is equivalent to finding the smallest prime factor of n, then dividing n by that factor

Let me trace through examples:
- n=15: factors are 1,3,5,15. Largest proper divisor = 5 (15/3)
- n=12: factors are 1,2,3,4,6,12. Largest proper divisor = 6 (12/2)  
- n=7: prime, factors are 1,7. Largest proper divisor = 1

Here's my approach:
1. Check for divisors from 2 up to sqrt(n)
2. When I find the first divisor d, then n/d is the largest proper divisor
3. If no divisor found, n is prime and answer is 1

```python
def largest_divisor(n: int) -> int:
    """ For a given number n, find the largest number that divides n evenly, smaller than n
    >>> largest_divisor(15)
    5
    """
    if n <= 1:
        return None
    
    # Find the smallest divisor >= 2
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return n // i
    
    # If no divisor found, n is prime, so largest proper divisor is 1
    return 1
```

Test output (stdout)

.                                                                        [100%]
1 passed in 0.01s

System prompt

You are an expert programmer. Solve the task exactly as specified. Output your solution as fenced code blocks using the required file name(s) and the exact function/type signatures requested. Prefer correctness; do not include prose outside code unless asked.