To solve this problem, we need to compute the sum of all proper divrs of a given number ( n ), where a proper divr is a number that divides ( n ) exactly and is less than ( n ) itself.

Approach

The key insight to efficiently compute the sum of proper divrs is to leverage the properties of divrs:

1. For any number ( n ), divrs come in pairs ((i, n/i)) where ( i \leq \sqrt{n} ).
2. We only need to check divrs up to ( \sqrt{n} ) to avoid redundant calculations.
3. Handle edge cases: if ( n \leq 1 ), there are no proper divrs, so the sum is 0.
4. Include 1 as a proper divr (since it divides all positive integers and is less than ( n ) for ( n >1 )).
5. For each divr ( i ) found, add both ( i ) and ( n/i ) to the sum (unless they are the same, to avoid double-counting perfect squares).

Solution Code

def sum_divrs(n):
    if n <= 1:
        return 0
    total = 1  # 1 is a proper divr for all n > 1
    sqrt_n = int(n ** 0.5)
    for i in range(2, sqrt_n + 1):
        if n % i == 0:
            total += i
            counterpart = n // i
            if counterpart != i:
                total += counterpart
    return total

Explanation

Let’s break down the solution:

1. Edge Case Handling: If ( n ) is 1 or less, return 0 since there are no proper divrs.
2. Initialization: Start with a sum of 1 because it is a proper divr for all ( n >1 ).
3. Loop Through Divrs: Iterate from 2 up to the integer square root of ( n ):
   • For each ( i ) that divides ( n ):
     • Add ( i ) to the sum.
     • Compute the counterpart divr ( n//i ). If it is different from ( i ), add it to the sum (to avoid double-counting for perfect squares like 4, where ( i=2 ) and ( n//i=2 )).
4. Return Result: The total sum of proper divrs is returned.

Examples

• For ( n=6 ): Proper divrs are 1,2,3 → sum is (1+2+3=6).
• For ( n=28 ): Proper divrs are1,2,4,7,14 → sum is (1+2+4+7+14=28).
• For ( n=12 ): Proper divrs are 1,2,3,4,6 → sum is (1+2+3+4+6=16).

This approach efficiently computes the sum with a time complexity of ( O(\sqrt{n}) ), which is optimal for this problem. It handles all positive integers correctly and avoids unnecessary calculations.

Answer: The function sum_divrs(n) as defined above. For example, if you call sum_divrs(6), it returns 6; sum_divrs(28) returns28, etc. The code is ready to use.

def sum_divrs(n):
    if n <= 1:
        return 0
    total = 1
    sqrt_n = int(n ** 0.5)
    for i in range(2, sqrt_n + 1):
        if n % i == 0:
            total += i
            counterpart = n // i
            if counterpart != i:
                total += counterpart
    return total

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