What are the divisors of 4740?

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, 79, 158, 237, 316, 395, 474, 790, 948, 1185, 1580, 2370, 4740

There is a total of 24 positive divisors.
The sum of these divisors is 13440.
The arithmetic mean is 560.

16 even divisors

2, 4, 6, 10, 12, 20, 30, 60, 158, 316, 474, 790, 948, 1580, 2370, 4740

8 odd divisors

1, 3, 5, 15, 79, 237, 395, 1185

How to compute the divisors of 4740?

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 4740 by each of the numbers from 1 to 4740 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)

4740 / 1 = 4740 (the remainder is 0, so 1 is a divisor of 4740)
4740 / 2 = 2370 (the remainder is 0, so 2 is a divisor of 4740)
4740 / 3 = 1580 (the remainder is 0, so 3 is a divisor of 4740)
...
4740 / 4739 = 1.0002110149821 (the remainder is 1, so 4739 is not a divisor of 4740)
4740 / 4740 = 1 (the remainder is 0, so 4740 is a divisor of 4740)

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 4740 (i.e. 68.847657912234). 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 $\frac{N}{d} = D$ )

4740 / 1 = 4740 (the remainder is 0, so 1 and 4740 are divisors of 4740)
4740 / 2 = 2370 (the remainder is 0, so 2 and 2370 are divisors of 4740)
4740 / 3 = 1580 (the remainder is 0, so 3 and 1580 are divisors of 4740)
...
4740 / 67 = 70.746268656716 (the remainder is 50, so 67 is not a divisor of 4740)
4740 / 68 = 69.705882352941 (the remainder is 48, so 68 is not a divisor of 4740)