Problem

A positive integer N is abundant if the sum of its proper divisors (all positive divisors except N itself) is strictly greater than N. Given N, return true if it is abundant and false otherwise.

Related terms: if the sum equals N the number is perfect; if the sum is less it is deficient.

Input / Output

• Input: integer N.
• Output: boolean.

Examples

Example 1 Input: N = 12 Output: true Explanation: Proper divisors are 1, 2, 3, 4, 6; sum = 16 > 12.

Example 2 Input: N = 28 Output: false Explanation: Proper divisors 1, 2, 4, 7, 14 sum to 28 exactly — 28 is perfect, not abundant.

Example 3 Input: N = 15 Output: false Explanation: 1 + 3 + 5 = 9 < 15.

Constraints

• 1 <= N <= 10^9

Approach

Enumerating all divisors up to N is O(N) — too slow for 10^9. Instead iterate only up to √N, because divisors come in pairs (i, N/i):

sum = 1                      // 1 is a proper divisor of every N > 1
for i in 2 .. floor(sqrt(N)):
    if N % i == 0:
        sum += i
        if i != N / i: sum += N / i   // avoid double-counting a perfect square root
return sum > N

Handle N = 1 separately (no proper divisors → not abundant).

Use 64-bit integers for the sum; for highly composite N near 10^9 the divisor sum can exceed 2^31.

Complexity

• Time: O(√N)
• Space: O(1)

Follow-ups

• Precompute abundance for all numbers 1..M using a sieve-like loop (for each i, add i to every multiple of i), giving O(M log M) for many queries.
• Classify a number as abundant, perfect or deficient in one function.
• The smallest odd abundant number is 945 — a fun check for your implementation.

Reference solution (Python)

from math import isqrt

def is_abundant(n):
    if n <= 1:
        return False
    total = 1
    for i in range(2, isqrt(n) + 1):
        if n % i == 0:
            total += i
            if i != n // i:
                total += n // i
    return total > n

Test cases

Input	Expected
12	true
28	false (perfect)
15	false
945	true (smallest odd abundant)