What are the divisors of 2450?

1, 2, 5, 7, 10, 14, 25, 35, 49, 50, 70, 98, 175, 245, 350, 490, 1225, 2450

There is a total of 18 positive divisors.
The sum of these divisors is 5301.
The arithmetic mean is 294.5.

9 even divisors

2, 10, 14, 50, 70, 98, 350, 490, 2450

9 odd divisors

1, 5, 7, 25, 35, 49, 175, 245, 1225

How to compute the divisors of 2450?

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 2450 by each of the numbers from 1 to 2450 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)

2450 / 1 = 2450 (the remainder is 0, so 1 is a divisor of 2450)
2450 / 2 = 1225 (the remainder is 0, so 2 is a divisor of 2450)
2450 / 3 = 816.66666666667 (the remainder is 2, so 3 is not a divisor of 2450)
...
2450 / 2449 = 1.0004083299306 (the remainder is 1, so 2449 is not a divisor of 2450)
2450 / 2450 = 1 (the remainder is 0, so 2450 is a divisor of 2450)

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 2450 (i.e. 49.497474683058). 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 $\frac{N}{d} = D$ )

2450 / 1 = 2450 (the remainder is 0, so 1 and 2450 are divisors of 2450)
2450 / 2 = 1225 (the remainder is 0, so 2 and 1225 are divisors of 2450)
2450 / 3 = 816.66666666667 (the remainder is 2, so 3 is not a divisor of 2450)
...
2450 / 48 = 51.041666666667 (the remainder is 2, so 48 is not a divisor of 2450)
2450 / 49 = 50 (the remainder is 0, so 49 and 50 are divisors of 2450)