What are the divisors of 1770?

1, 2, 3, 5, 6, 10, 15, 30, 59, 118, 177, 295, 354, 590, 885, 1770

There is a total of 16 positive divisors.
The sum of these divisors is 4320.
The arithmetic mean is 270.

8 even divisors

2, 6, 10, 30, 118, 354, 590, 1770

8 odd divisors

1, 3, 5, 15, 59, 177, 295, 885

How to compute the divisors of 1770?

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 1770 by each of the numbers from 1 to 1770 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)

1770 / 1 = 1770 (the remainder is 0, so 1 is a divisor of 1770)
1770 / 2 = 885 (the remainder is 0, so 2 is a divisor of 1770)
1770 / 3 = 590 (the remainder is 0, so 3 is a divisor of 1770)
...
1770 / 1769 = 1.0005652911249 (the remainder is 1, so 1769 is not a divisor of 1770)
1770 / 1770 = 1 (the remainder is 0, so 1770 is a divisor of 1770)

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 1770 (i.e. 42.071367935925). 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 $\frac{N}{d} = D$ )

1770 / 1 = 1770 (the remainder is 0, so 1 and 1770 are divisors of 1770)
1770 / 2 = 885 (the remainder is 0, so 2 and 885 are divisors of 1770)
1770 / 3 = 590 (the remainder is 0, so 3 and 590 are divisors of 1770)
...
1770 / 41 = 43.170731707317 (the remainder is 7, so 41 is not a divisor of 1770)
1770 / 42 = 42.142857142857 (the remainder is 6, so 42 is not a divisor of 1770)