What are the divisors of 3447?

1, 3, 9, 383, 1149, 3447

There is a total of 6 positive divisors.
The sum of these divisors is 4992.
The arithmetic mean is 832.

6 odd divisors

1, 3, 9, 383, 1149, 3447

How to compute the divisors of 3447?

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 3447 by each of the numbers from 1 to 3447 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)

3447 / 1 = 3447 (the remainder is 0, so 1 is a divisor of 3447)
3447 / 2 = 1723.5 (the remainder is 1, so 2 is not a divisor of 3447)
3447 / 3 = 1149 (the remainder is 0, so 3 is a divisor of 3447)
...
3447 / 3446 = 1.0002901915264 (the remainder is 1, so 3446 is not a divisor of 3447)
3447 / 3447 = 1 (the remainder is 0, so 3447 is a divisor of 3447)

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 3447 (i.e. 58.711157372343). 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 $\frac{N}{d} = D$ )

3447 / 1 = 3447 (the remainder is 0, so 1 and 3447 are divisors of 3447)
3447 / 2 = 1723.5 (the remainder is 1, so 2 is not a divisor of 3447)
3447 / 3 = 1149 (the remainder is 0, so 3 and 1149 are divisors of 3447)
...
3447 / 57 = 60.473684210526 (the remainder is 27, so 57 is not a divisor of 3447)
3447 / 58 = 59.431034482759 (the remainder is 25, so 58 is not a divisor of 3447)