What are the divisors of 3248?

1, 2, 4, 7, 8, 14, 16, 28, 29, 56, 58, 112, 116, 203, 232, 406, 464, 812, 1624, 3248

There is a total of 20 positive divisors.
The sum of these divisors is 7440.
The arithmetic mean is 372.

16 even divisors

2, 4, 8, 14, 16, 28, 56, 58, 112, 116, 232, 406, 464, 812, 1624, 3248

4 odd divisors

1, 7, 29, 203

How to compute the divisors of 3248?

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 3248 by each of the numbers from 1 to 3248 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)

3248 / 1 = 3248 (the remainder is 0, so 1 is a divisor of 3248)
3248 / 2 = 1624 (the remainder is 0, so 2 is a divisor of 3248)
3248 / 3 = 1082.6666666667 (the remainder is 2, so 3 is not a divisor of 3248)
...
3248 / 3247 = 1.0003079765938 (the remainder is 1, so 3247 is not a divisor of 3248)
3248 / 3248 = 1 (the remainder is 0, so 3248 is a divisor of 3248)

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 3248 (i.e. 56.9912273951). 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 $\frac{N}{d} = D$ )

3248 / 1 = 3248 (the remainder is 0, so 1 and 3248 are divisors of 3248)
3248 / 2 = 1624 (the remainder is 0, so 2 and 1624 are divisors of 3248)
3248 / 3 = 1082.6666666667 (the remainder is 2, so 3 is not a divisor of 3248)
...
3248 / 55 = 59.054545454545 (the remainder is 3, so 55 is not a divisor of 3248)
3248 / 56 = 58 (the remainder is 0, so 56 and 58 are divisors of 3248)