What are the divisors of 802?

1, 2, 401, 802

There is a total of 4 positive divisors.
The sum of these divisors is 1206.
The arithmetic mean is 301.5.

2 even divisors

2, 802

2 odd divisors

1, 401

How to compute the divisors of 802?

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 802 by each of the numbers from 1 to 802 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)

802 / 1 = 802 (the remainder is 0, so 1 is a divisor of 802)
802 / 2 = 401 (the remainder is 0, so 2 is a divisor of 802)
802 / 3 = 267.33333333333 (the remainder is 1, so 3 is not a divisor of 802)
...
802 / 801 = 1.0012484394507 (the remainder is 1, so 801 is not a divisor of 802)
802 / 802 = 1 (the remainder is 0, so 802 is a divisor of 802)

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 802 (i.e. 28.319604517013). 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 $\frac{N}{d} = D$ )

802 / 1 = 802 (the remainder is 0, so 1 and 802 are divisors of 802)
802 / 2 = 401 (the remainder is 0, so 2 and 401 are divisors of 802)
802 / 3 = 267.33333333333 (the remainder is 1, so 3 is not a divisor of 802)
...
802 / 27 = 29.703703703704 (the remainder is 19, so 27 is not a divisor of 802)
802 / 28 = 28.642857142857 (the remainder is 18, so 28 is not a divisor of 802)