What are the divisors of 3718?

1, 2, 11, 13, 22, 26, 143, 169, 286, 338, 1859, 3718

There is a total of 12 positive divisors.
The sum of these divisors is 6588.
The arithmetic mean is 549.

6 even divisors

2, 22, 26, 286, 338, 3718

6 odd divisors

1, 11, 13, 143, 169, 1859

How to compute the divisors of 3718?

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 3718 by each of the numbers from 1 to 3718 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)

3718 / 1 = 3718 (the remainder is 0, so 1 is a divisor of 3718)
3718 / 2 = 1859 (the remainder is 0, so 2 is a divisor of 3718)
3718 / 3 = 1239.3333333333 (the remainder is 1, so 3 is not a divisor of 3718)
...
3718 / 3717 = 1.0002690341673 (the remainder is 1, so 3717 is not a divisor of 3718)
3718 / 3718 = 1 (the remainder is 0, so 3718 is a divisor of 3718)

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 3718 (i.e. 60.975404877705). 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 $\frac{N}{d} = D$ )

3718 / 1 = 3718 (the remainder is 0, so 1 and 3718 are divisors of 3718)
3718 / 2 = 1859 (the remainder is 0, so 2 and 1859 are divisors of 3718)
3718 / 3 = 1239.3333333333 (the remainder is 1, so 3 is not a divisor of 3718)
...
3718 / 59 = 63.016949152542 (the remainder is 1, so 59 is not a divisor of 3718)
3718 / 60 = 61.966666666667 (the remainder is 58, so 60 is not a divisor of 3718)