What are the divisors of 3948?

1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 47, 84, 94, 141, 188, 282, 329, 564, 658, 987, 1316, 1974, 3948

There is a total of 24 positive divisors.
The sum of these divisors is 10752.
The arithmetic mean is 448.

16 even divisors

2, 4, 6, 12, 14, 28, 42, 84, 94, 188, 282, 564, 658, 1316, 1974, 3948

8 odd divisors

1, 3, 7, 21, 47, 141, 329, 987

How to compute the divisors of 3948?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 3948 by each of the numbers from 1 to 3948 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

3948 / 1 = 3948 (the remainder is 0, so 1 is a divisor of 3948)
3948 / 2 = 1974 (the remainder is 0, so 2 is a divisor of 3948)
3948 / 3 = 1316 (the remainder is 0, so 3 is a divisor of 3948)
...
3948 / 3947 = 1.00025335698 (the remainder is 1, so 3947 is not a divisor of 3948)
3948 / 3948 = 1 (the remainder is 0, so 3948 is a divisor of 3948)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 3948 (i.e. 62.83311228962). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

3948 / 1 = 3948 (the remainder is 0, so 1 and 3948 are divisors of 3948)
3948 / 2 = 1974 (the remainder is 0, so 2 and 1974 are divisors of 3948)
3948 / 3 = 1316 (the remainder is 0, so 3 and 1316 are divisors of 3948)
...
3948 / 61 = 64.72131147541 (the remainder is 44, so 61 is not a divisor of 3948)
3948 / 62 = 63.677419354839 (the remainder is 42, so 62 is not a divisor of 3948)