What are the divisors of 1210?

1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605, 1210

There is a total of 12 positive divisors.
The sum of these divisors is 2394.
The arithmetic mean is 199.5.

6 even divisors

2, 10, 22, 110, 242, 1210

6 odd divisors

1, 5, 11, 55, 121, 605

How to compute the divisors of 1210?

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 1210 by each of the numbers from 1 to 1210 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)

1210 / 1 = 1210 (the remainder is 0, so 1 is a divisor of 1210)
1210 / 2 = 605 (the remainder is 0, so 2 is a divisor of 1210)
1210 / 3 = 403.33333333333 (the remainder is 1, so 3 is not a divisor of 1210)
...
1210 / 1209 = 1.0008271298594 (the remainder is 1, so 1209 is not a divisor of 1210)
1210 / 1210 = 1 (the remainder is 0, so 1210 is a divisor of 1210)

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 1210 (i.e. 34.785054261852). 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 $\frac{N}{d} = D$ )

1210 / 1 = 1210 (the remainder is 0, so 1 and 1210 are divisors of 1210)
1210 / 2 = 605 (the remainder is 0, so 2 and 605 are divisors of 1210)
1210 / 3 = 403.33333333333 (the remainder is 1, so 3 is not a divisor of 1210)
...
1210 / 33 = 36.666666666667 (the remainder is 22, so 33 is not a divisor of 1210)
1210 / 34 = 35.588235294118 (the remainder is 20, so 34 is not a divisor of 1210)