What are the divisors of 202?

1, 2, 101, 202

There is a total of 4 positive divisors.
The sum of these divisors is 306.
The arithmetic mean is 76.5.

2 even divisors

2, 202

2 odd divisors

1, 101

How to compute the divisors of 202?

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 202 by each of the numbers from 1 to 202 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)

202 / 1 = 202 (the remainder is 0, so 1 is a divisor of 202)
202 / 2 = 101 (the remainder is 0, so 2 is a divisor of 202)
202 / 3 = 67.333333333333 (the remainder is 1, so 3 is not a divisor of 202)
...
202 / 201 = 1.0049751243781 (the remainder is 1, so 201 is not a divisor of 202)
202 / 202 = 1 (the remainder is 0, so 202 is a divisor of 202)

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 202 (i.e. 14.212670403552). 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 $\frac{N}{d} = D$ )

202 / 1 = 202 (the remainder is 0, so 1 and 202 are divisors of 202)
202 / 2 = 101 (the remainder is 0, so 2 and 101 are divisors of 202)
202 / 3 = 67.333333333333 (the remainder is 1, so 3 is not a divisor of 202)
...
202 / 13 = 15.538461538462 (the remainder is 7, so 13 is not a divisor of 202)
202 / 14 = 14.428571428571 (the remainder is 6, so 14 is not a divisor of 202)