What are the divisors of 3950?

1, 2, 5, 10, 25, 50, 79, 158, 395, 790, 1975, 3950

There is a total of 12 positive divisors.
The sum of these divisors is 7440.
The arithmetic mean is 620.

6 even divisors

2, 10, 50, 158, 790, 3950

6 odd divisors

1, 5, 25, 79, 395, 1975

How to compute the divisors of 3950?

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 3950 by each of the numbers from 1 to 3950 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)

3950 / 1 = 3950 (the remainder is 0, so 1 is a divisor of 3950)
3950 / 2 = 1975 (the remainder is 0, so 2 is a divisor of 3950)
3950 / 3 = 1316.6666666667 (the remainder is 2, so 3 is not a divisor of 3950)
...
3950 / 3949 = 1.0002532286655 (the remainder is 1, so 3949 is not a divisor of 3950)
3950 / 3950 = 1 (the remainder is 0, so 3950 is a divisor of 3950)

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 3950 (i.e. 62.849025449883). 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 $\frac{N}{d} = D$ )

3950 / 1 = 3950 (the remainder is 0, so 1 and 3950 are divisors of 3950)
3950 / 2 = 1975 (the remainder is 0, so 2 and 1975 are divisors of 3950)
3950 / 3 = 1316.6666666667 (the remainder is 2, so 3 is not a divisor of 3950)
...
3950 / 61 = 64.754098360656 (the remainder is 46, so 61 is not a divisor of 3950)
3950 / 62 = 63.709677419355 (the remainder is 44, so 62 is not a divisor of 3950)