To solve this problem (assuming it refers to Project Euler Problem 21: Amicable Numbers), we need to find the sum of all amicable numbers under 10000. Amicable numbers are pairs where each number is the sum of the proper divrs of the other.

Approach

1. Sum of Proper Divrs: For each number n, compute the sum of its proper divrs (divrs excluding n itself). This can be done efficiently by iterating up to the square root of n and adding both divrs if they are distinct.
2. Identify Amicable Pairs: For each number a from 2 to 9999:
   • Compute b as the sum of proper divrs of a.
   • Check if b > a (to avoid duplicate pairs) and if the sum of proper divrs of b equals a.
   • If both conditions are met, add both a and b to the total sum.

Solution Code

def sum_proper_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
            other = n // i
            if other != i:
                total += other
    return total

sum_amicable = 0
seen = set()

for a in range(2, 10000):
    if a in seen:
        continue
    b = sum_proper_divrs(a)
    if b > a and b < 10000 and sum_proper_divrs(b) == a:
        sum_amicable += a + b
        seen.add(a)
        seen.add(b)

print(sum_amicable)

Explanation

• Sum of Proper Divrs: The function sum_proper_divrs calculates the sum of divrs by checking up to the square root of n, ensuring efficiency (O(√n) per number).
• Amicable Pairs: By checking b > a, we avoid processing the same pair twice (e.g., 220 and 284 are only considered once when a=220). The seen set ensures we don't add duplicate numbers to the sum.

Answer: 31626. This is the sum of all amicable numbers under 10000.

\boxed{31626}

作者声明:本文包含人工智能生成内容。