What are the divisors of 3202?

1, 2, 1601, 3202

There is a total of 4 positive divisors.
The sum of these divisors is 4806.
The arithmetic mean is 1201.5.

2 even divisors

2, 3202

2 odd divisors

1, 1601

How to compute the divisors of 3202?

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 3202 by each of the numbers from 1 to 3202 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)

3202 / 1 = 3202 (the remainder is 0, so 1 is a divisor of 3202)
3202 / 2 = 1601 (the remainder is 0, so 2 is a divisor of 3202)
3202 / 3 = 1067.3333333333 (the remainder is 1, so 3 is not a divisor of 3202)
...
3202 / 3201 = 1.0003124023743 (the remainder is 1, so 3201 is not a divisor of 3202)
3202 / 3202 = 1 (the remainder is 0, so 3202 is a divisor of 3202)

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 3202 (i.e. 56.58621740318). 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 $\frac{N}{d} = D$ )

3202 / 1 = 3202 (the remainder is 0, so 1 and 3202 are divisors of 3202)
3202 / 2 = 1601 (the remainder is 0, so 2 and 1601 are divisors of 3202)
3202 / 3 = 1067.3333333333 (the remainder is 1, so 3 is not a divisor of 3202)
...
3202 / 55 = 58.218181818182 (the remainder is 12, so 55 is not a divisor of 3202)
3202 / 56 = 57.178571428571 (the remainder is 10, so 56 is not a divisor of 3202)