What are the divisors of 1238?

1, 2, 619, 1238

There is a total of 4 positive divisors.
The sum of these divisors is 1860.
The arithmetic mean is 465.

2 even divisors

2, 1238

2 odd divisors

1, 619

How to compute the divisors of 1238?

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 1238 by each of the numbers from 1 to 1238 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)

1238 / 1 = 1238 (the remainder is 0, so 1 is a divisor of 1238)
1238 / 2 = 619 (the remainder is 0, so 2 is a divisor of 1238)
1238 / 3 = 412.66666666667 (the remainder is 2, so 3 is not a divisor of 1238)
...
1238 / 1237 = 1.0008084074373 (the remainder is 1, so 1237 is not a divisor of 1238)
1238 / 1238 = 1 (the remainder is 0, so 1238 is a divisor of 1238)

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 1238 (i.e. 35.185224171518). 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 $\frac{N}{d} = D$ )

1238 / 1 = 1238 (the remainder is 0, so 1 and 1238 are divisors of 1238)
1238 / 2 = 619 (the remainder is 0, so 2 and 619 are divisors of 1238)
1238 / 3 = 412.66666666667 (the remainder is 2, so 3 is not a divisor of 1238)
...
1238 / 34 = 36.411764705882 (the remainder is 14, so 34 is not a divisor of 1238)
1238 / 35 = 35.371428571429 (the remainder is 13, so 35 is not a divisor of 1238)