What are the divisors of 3235?

1, 5, 647, 3235

There is a total of 4 positive divisors.
The sum of these divisors is 3888.
The arithmetic mean is 972.

4 odd divisors

1, 5, 647, 3235

How to compute the divisors of 3235?

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 3235 by each of the numbers from 1 to 3235 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)

3235 / 1 = 3235 (the remainder is 0, so 1 is a divisor of 3235)
3235 / 2 = 1617.5 (the remainder is 1, so 2 is not a divisor of 3235)
3235 / 3 = 1078.3333333333 (the remainder is 1, so 3 is not a divisor of 3235)
...
3235 / 3234 = 1.0003092145949 (the remainder is 1, so 3234 is not a divisor of 3235)
3235 / 3235 = 1 (the remainder is 0, so 3235 is a divisor of 3235)

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 3235 (i.e. 56.877060402239). 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 $\frac{N}{d} = D$ )

3235 / 1 = 3235 (the remainder is 0, so 1 and 3235 are divisors of 3235)
3235 / 2 = 1617.5 (the remainder is 1, so 2 is not a divisor of 3235)
3235 / 3 = 1078.3333333333 (the remainder is 1, so 3 is not a divisor of 3235)
...
3235 / 55 = 58.818181818182 (the remainder is 45, so 55 is not a divisor of 3235)
3235 / 56 = 57.767857142857 (the remainder is 43, so 56 is not a divisor of 3235)