What are the divisors of 2903?

1, 2903

There is a total of 2 positive divisors.
The sum of these divisors is 2904.
The arithmetic mean is 1452.

2 odd divisors

1, 2903

How to compute the divisors of 2903?

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 2903 by each of the numbers from 1 to 2903 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)

2903 / 1 = 2903 (the remainder is 0, so 1 is a divisor of 2903)
2903 / 2 = 1451.5 (the remainder is 1, so 2 is not a divisor of 2903)
2903 / 3 = 967.66666666667 (the remainder is 2, so 3 is not a divisor of 2903)
...
2903 / 2902 = 1.000344589938 (the remainder is 1, so 2902 is not a divisor of 2903)
2903 / 2903 = 1 (the remainder is 0, so 2903 is a divisor of 2903)

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 2903 (i.e. 53.879495172097). 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 $\frac{N}{d} = D$ )

2903 / 1 = 2903 (the remainder is 0, so 1 and 2903 are divisors of 2903)
2903 / 2 = 1451.5 (the remainder is 1, so 2 is not a divisor of 2903)
2903 / 3 = 967.66666666667 (the remainder is 2, so 3 is not a divisor of 2903)
...
2903 / 52 = 55.826923076923 (the remainder is 43, so 52 is not a divisor of 2903)
2903 / 53 = 54.77358490566 (the remainder is 41, so 53 is not a divisor of 2903)