What are the divisors of 1703?

1, 13, 131, 1703

There is a total of 4 positive divisors.
The sum of these divisors is 1848.
The arithmetic mean is 462.

4 odd divisors

1, 13, 131, 1703

How to compute the divisors of 1703?

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 1703 by each of the numbers from 1 to 1703 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)

1703 / 1 = 1703 (the remainder is 0, so 1 is a divisor of 1703)
1703 / 2 = 851.5 (the remainder is 1, so 2 is not a divisor of 1703)
1703 / 3 = 567.66666666667 (the remainder is 2, so 3 is not a divisor of 1703)
...
1703 / 1702 = 1.0005875440658 (the remainder is 1, so 1702 is not a divisor of 1703)
1703 / 1703 = 1 (the remainder is 0, so 1703 is a divisor of 1703)

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 1703 (i.e. 41.267420563927). 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 $\frac{N}{d} = D$ )

1703 / 1 = 1703 (the remainder is 0, so 1 and 1703 are divisors of 1703)
1703 / 2 = 851.5 (the remainder is 1, so 2 is not a divisor of 1703)
1703 / 3 = 567.66666666667 (the remainder is 2, so 3 is not a divisor of 1703)
...
1703 / 40 = 42.575 (the remainder is 23, so 40 is not a divisor of 1703)
1703 / 41 = 41.536585365854 (the remainder is 22, so 41 is not a divisor of 1703)