What are the divisors of 2902?

1, 2, 1451, 2902

There is a total of 4 positive divisors.
The sum of these divisors is 4356.
The arithmetic mean is 1089.

2 even divisors

2, 2902

2 odd divisors

1, 1451

How to compute the divisors of 2902?

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 2902 by each of the numbers from 1 to 2902 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)

2902 / 1 = 2902 (the remainder is 0, so 1 is a divisor of 2902)
2902 / 2 = 1451 (the remainder is 0, so 2 is a divisor of 2902)
2902 / 3 = 967.33333333333 (the remainder is 1, so 3 is not a divisor of 2902)
...
2902 / 2901 = 1.0003447087211 (the remainder is 1, so 2901 is not a divisor of 2902)
2902 / 2902 = 1 (the remainder is 0, so 2902 is a divisor of 2902)

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 2902 (i.e. 53.870214404623). 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 $\frac{N}{d} = D$ )

2902 / 1 = 2902 (the remainder is 0, so 1 and 2902 are divisors of 2902)
2902 / 2 = 1451 (the remainder is 0, so 2 and 1451 are divisors of 2902)
2902 / 3 = 967.33333333333 (the remainder is 1, so 3 is not a divisor of 2902)
...
2902 / 52 = 55.807692307692 (the remainder is 42, so 52 is not a divisor of 2902)
2902 / 53 = 54.754716981132 (the remainder is 40, so 53 is not a divisor of 2902)