What are the divisors of 2975?

1, 5, 7, 17, 25, 35, 85, 119, 175, 425, 595, 2975

There is a total of 12 positive divisors.
The sum of these divisors is 4464.
The arithmetic mean is 372.

12 odd divisors

1, 5, 7, 17, 25, 35, 85, 119, 175, 425, 595, 2975

How to compute the divisors of 2975?

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 2975 by each of the numbers from 1 to 2975 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)

2975 / 1 = 2975 (the remainder is 0, so 1 is a divisor of 2975)
2975 / 2 = 1487.5 (the remainder is 1, so 2 is not a divisor of 2975)
2975 / 3 = 991.66666666667 (the remainder is 2, so 3 is not a divisor of 2975)
...
2975 / 2974 = 1.0003362474781 (the remainder is 1, so 2974 is not a divisor of 2975)
2975 / 2975 = 1 (the remainder is 0, so 2975 is a divisor of 2975)

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 2975 (i.e. 54.543560573179). 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 $\frac{N}{d} = D$ )

2975 / 1 = 2975 (the remainder is 0, so 1 and 2975 are divisors of 2975)
2975 / 2 = 1487.5 (the remainder is 1, so 2 is not a divisor of 2975)
2975 / 3 = 991.66666666667 (the remainder is 2, so 3 is not a divisor of 2975)
...
2975 / 53 = 56.132075471698 (the remainder is 7, so 53 is not a divisor of 2975)
2975 / 54 = 55.092592592593 (the remainder is 5, so 54 is not a divisor of 2975)