What are the divisors of 3525?

1, 3, 5, 15, 25, 47, 75, 141, 235, 705, 1175, 3525

There is a total of 12 positive divisors.
The sum of these divisors is 5952.
The arithmetic mean is 496.

12 odd divisors

1, 3, 5, 15, 25, 47, 75, 141, 235, 705, 1175, 3525

How to compute the divisors of 3525?

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 3525 by each of the numbers from 1 to 3525 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)

3525 / 1 = 3525 (the remainder is 0, so 1 is a divisor of 3525)
3525 / 2 = 1762.5 (the remainder is 1, so 2 is not a divisor of 3525)
3525 / 3 = 1175 (the remainder is 0, so 3 is a divisor of 3525)
...
3525 / 3524 = 1.0002837684449 (the remainder is 1, so 3524 is not a divisor of 3525)
3525 / 3525 = 1 (the remainder is 0, so 3525 is a divisor of 3525)

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 3525 (i.e. 59.37171043519). 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 $\frac{N}{d} = D$ )

3525 / 1 = 3525 (the remainder is 0, so 1 and 3525 are divisors of 3525)
3525 / 2 = 1762.5 (the remainder is 1, so 2 is not a divisor of 3525)
3525 / 3 = 1175 (the remainder is 0, so 3 and 1175 are divisors of 3525)
...
3525 / 58 = 60.775862068966 (the remainder is 45, so 58 is not a divisor of 3525)
3525 / 59 = 59.745762711864 (the remainder is 44, so 59 is not a divisor of 3525)