What are the divisors of 2602?

1, 2, 1301, 2602

There is a total of 4 positive divisors.
The sum of these divisors is 3906.
The arithmetic mean is 976.5.

2 even divisors

2, 2602

2 odd divisors

1, 1301

How to compute the divisors of 2602?

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 2602 by each of the numbers from 1 to 2602 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)

2602 / 1 = 2602 (the remainder is 0, so 1 is a divisor of 2602)
2602 / 2 = 1301 (the remainder is 0, so 2 is a divisor of 2602)
2602 / 3 = 867.33333333333 (the remainder is 1, so 3 is not a divisor of 2602)
...
2602 / 2601 = 1.0003844675125 (the remainder is 1, so 2601 is not a divisor of 2602)
2602 / 2602 = 1 (the remainder is 0, so 2602 is a divisor of 2602)

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 2602 (i.e. 51.009802979427). 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 $\frac{N}{d} = D$ )

2602 / 1 = 2602 (the remainder is 0, so 1 and 2602 are divisors of 2602)
2602 / 2 = 1301 (the remainder is 0, so 2 and 1301 are divisors of 2602)
2602 / 3 = 867.33333333333 (the remainder is 1, so 3 is not a divisor of 2602)
...
2602 / 50 = 52.04 (the remainder is 2, so 50 is not a divisor of 2602)
2602 / 51 = 51.019607843137 (the remainder is 1, so 51 is not a divisor of 2602)