What are the divisors of 1975?

1, 5, 25, 79, 395, 1975

There is a total of 6 positive divisors.
The sum of these divisors is 2480.
The arithmetic mean is 413.33333333333.

6 odd divisors

1, 5, 25, 79, 395, 1975

How to compute the divisors of 1975?

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 1975 by each of the numbers from 1 to 1975 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)

1975 / 1 = 1975 (the remainder is 0, so 1 is a divisor of 1975)
1975 / 2 = 987.5 (the remainder is 1, so 2 is not a divisor of 1975)
1975 / 3 = 658.33333333333 (the remainder is 1, so 3 is not a divisor of 1975)
...
1975 / 1974 = 1.000506585613 (the remainder is 1, so 1974 is not a divisor of 1975)
1975 / 1975 = 1 (the remainder is 0, so 1975 is a divisor of 1975)

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 1975 (i.e. 44.440972086578). 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 $\frac{N}{d} = D$ )

1975 / 1 = 1975 (the remainder is 0, so 1 and 1975 are divisors of 1975)
1975 / 2 = 987.5 (the remainder is 1, so 2 is not a divisor of 1975)
1975 / 3 = 658.33333333333 (the remainder is 1, so 3 is not a divisor of 1975)
...
1975 / 43 = 45.93023255814 (the remainder is 40, so 43 is not a divisor of 1975)
1975 / 44 = 44.886363636364 (the remainder is 39, so 44 is not a divisor of 1975)