What are the divisors of 3762?

1, 2, 3, 6, 9, 11, 18, 19, 22, 33, 38, 57, 66, 99, 114, 171, 198, 209, 342, 418, 627, 1254, 1881, 3762

There is a total of 24 positive divisors.
The sum of these divisors is 9360.
The arithmetic mean is 390.

12 even divisors

2, 6, 18, 22, 38, 66, 114, 198, 342, 418, 1254, 3762

12 odd divisors

1, 3, 9, 11, 19, 33, 57, 99, 171, 209, 627, 1881

How to compute the divisors of 3762?

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 3762 by each of the numbers from 1 to 3762 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)

3762 / 1 = 3762 (the remainder is 0, so 1 is a divisor of 3762)
3762 / 2 = 1881 (the remainder is 0, so 2 is a divisor of 3762)
3762 / 3 = 1254 (the remainder is 0, so 3 is a divisor of 3762)
...
3762 / 3761 = 1.0002658867323 (the remainder is 1, so 3761 is not a divisor of 3762)
3762 / 3762 = 1 (the remainder is 0, so 3762 is a divisor of 3762)

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 3762 (i.e. 61.335144900783). 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 $\frac{N}{d} = D$ )

3762 / 1 = 3762 (the remainder is 0, so 1 and 3762 are divisors of 3762)
3762 / 2 = 1881 (the remainder is 0, so 2 and 1881 are divisors of 3762)
3762 / 3 = 1254 (the remainder is 0, so 3 and 1254 are divisors of 3762)
...
3762 / 60 = 62.7 (the remainder is 42, so 60 is not a divisor of 3762)
3762 / 61 = 61.672131147541 (the remainder is 41, so 61 is not a divisor of 3762)