What are the divisors of 7003?

1, 47, 149, 7003

There is a total of 4 positive divisors.
The sum of these divisors is 7200.
The arithmetic mean is 1800.

4 odd divisors

1, 47, 149, 7003

How to compute the divisors of 7003?

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 7003 by each of the numbers from 1 to 7003 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)

7003 / 1 = 7003 (the remainder is 0, so 1 is a divisor of 7003)
7003 / 2 = 3501.5 (the remainder is 1, so 2 is not a divisor of 7003)
7003 / 3 = 2334.3333333333 (the remainder is 1, so 3 is not a divisor of 7003)
...
7003 / 7002 = 1.0001428163382 (the remainder is 1, so 7002 is not a divisor of 7003)
7003 / 7003 = 1 (the remainder is 0, so 7003 is a divisor of 7003)

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 7003 (i.e. 83.683929162056). 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 $\frac{N}{d} = D$ )

7003 / 1 = 7003 (the remainder is 0, so 1 and 7003 are divisors of 7003)
7003 / 2 = 3501.5 (the remainder is 1, so 2 is not a divisor of 7003)
7003 / 3 = 2334.3333333333 (the remainder is 1, so 3 is not a divisor of 7003)
...
7003 / 82 = 85.40243902439 (the remainder is 33, so 82 is not a divisor of 7003)
7003 / 83 = 84.373493975904 (the remainder is 31, so 83 is not a divisor of 7003)