What are the divisors of 1748?

1, 2, 4, 19, 23, 38, 46, 76, 92, 437, 874, 1748

There is a total of 12 positive divisors.
The sum of these divisors is 3360.
The arithmetic mean is 280.

8 even divisors

2, 4, 38, 46, 76, 92, 874, 1748

4 odd divisors

1, 19, 23, 437

How to compute the divisors of 1748?

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 1748 by each of the numbers from 1 to 1748 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)

1748 / 1 = 1748 (the remainder is 0, so 1 is a divisor of 1748)
1748 / 2 = 874 (the remainder is 0, so 2 is a divisor of 1748)
1748 / 3 = 582.66666666667 (the remainder is 2, so 3 is not a divisor of 1748)
...
1748 / 1747 = 1.0005724098454 (the remainder is 1, so 1747 is not a divisor of 1748)
1748 / 1748 = 1 (the remainder is 0, so 1748 is a divisor of 1748)

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 1748 (i.e. 41.809089920734). 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 $\frac{N}{d} = D$ )

1748 / 1 = 1748 (the remainder is 0, so 1 and 1748 are divisors of 1748)
1748 / 2 = 874 (the remainder is 0, so 2 and 874 are divisors of 1748)
1748 / 3 = 582.66666666667 (the remainder is 2, so 3 is not a divisor of 1748)
...
1748 / 40 = 43.7 (the remainder is 28, so 40 is not a divisor of 1748)
1748 / 41 = 42.634146341463 (the remainder is 26, so 41 is not a divisor of 1748)