What are the divisors of 1253?

1, 7, 179, 1253

There is a total of 4 positive divisors.
The sum of these divisors is 1440.
The arithmetic mean is 360.

4 odd divisors

1, 7, 179, 1253

How to compute the divisors of 1253?

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 1253 by each of the numbers from 1 to 1253 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)

1253 / 1 = 1253 (the remainder is 0, so 1 is a divisor of 1253)
1253 / 2 = 626.5 (the remainder is 1, so 2 is not a divisor of 1253)
1253 / 3 = 417.66666666667 (the remainder is 2, so 3 is not a divisor of 1253)
...
1253 / 1252 = 1.0007987220447 (the remainder is 1, so 1252 is not a divisor of 1253)
1253 / 1253 = 1 (the remainder is 0, so 1253 is a divisor of 1253)

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 1253 (i.e. 35.397740040856). 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 $\frac{N}{d} = D$ )

1253 / 1 = 1253 (the remainder is 0, so 1 and 1253 are divisors of 1253)
1253 / 2 = 626.5 (the remainder is 1, so 2 is not a divisor of 1253)
1253 / 3 = 417.66666666667 (the remainder is 2, so 3 is not a divisor of 1253)
...
1253 / 34 = 36.852941176471 (the remainder is 29, so 34 is not a divisor of 1253)
1253 / 35 = 35.8 (the remainder is 28, so 35 is not a divisor of 1253)