What are the divisors of 1290?

1, 2, 3, 5, 6, 10, 15, 30, 43, 86, 129, 215, 258, 430, 645, 1290

There is a total of 16 positive divisors.
The sum of these divisors is 3168.
The arithmetic mean is 198.

8 even divisors

2, 6, 10, 30, 86, 258, 430, 1290

8 odd divisors

1, 3, 5, 15, 43, 129, 215, 645

How to compute the divisors of 1290?

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 1290 by each of the numbers from 1 to 1290 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)

1290 / 1 = 1290 (the remainder is 0, so 1 is a divisor of 1290)
1290 / 2 = 645 (the remainder is 0, so 2 is a divisor of 1290)
1290 / 3 = 430 (the remainder is 0, so 3 is a divisor of 1290)
...
1290 / 1289 = 1.0007757951901 (the remainder is 1, so 1289 is not a divisor of 1290)
1290 / 1290 = 1 (the remainder is 0, so 1290 is a divisor of 1290)

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 1290 (i.e. 35.916569992136). 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 $\frac{N}{d} = D$ )

1290 / 1 = 1290 (the remainder is 0, so 1 and 1290 are divisors of 1290)
1290 / 2 = 645 (the remainder is 0, so 2 and 645 are divisors of 1290)
1290 / 3 = 430 (the remainder is 0, so 3 and 430 are divisors of 1290)
...
1290 / 34 = 37.941176470588 (the remainder is 32, so 34 is not a divisor of 1290)
1290 / 35 = 36.857142857143 (the remainder is 30, so 35 is not a divisor of 1290)