What are the divisors of 3498?

1, 2, 3, 6, 11, 22, 33, 53, 66, 106, 159, 318, 583, 1166, 1749, 3498

There is a total of 16 positive divisors.
The sum of these divisors is 7776.
The arithmetic mean is 486.

8 even divisors

2, 6, 22, 66, 106, 318, 1166, 3498

8 odd divisors

1, 3, 11, 33, 53, 159, 583, 1749

How to compute the divisors of 3498?

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 3498 by each of the numbers from 1 to 3498 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)

3498 / 1 = 3498 (the remainder is 0, so 1 is a divisor of 3498)
3498 / 2 = 1749 (the remainder is 0, so 2 is a divisor of 3498)
3498 / 3 = 1166 (the remainder is 0, so 3 is a divisor of 3498)
...
3498 / 3497 = 1.0002859593938 (the remainder is 1, so 3497 is not a divisor of 3498)
3498 / 3498 = 1 (the remainder is 0, so 3498 is a divisor of 3498)

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 3498 (i.e. 59.143892330485). 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 $\frac{N}{d} = D$ )

3498 / 1 = 3498 (the remainder is 0, so 1 and 3498 are divisors of 3498)
3498 / 2 = 1749 (the remainder is 0, so 2 and 1749 are divisors of 3498)
3498 / 3 = 1166 (the remainder is 0, so 3 and 1166 are divisors of 3498)
...
3498 / 58 = 60.310344827586 (the remainder is 18, so 58 is not a divisor of 3498)
3498 / 59 = 59.28813559322 (the remainder is 17, so 59 is not a divisor of 3498)