What are the divisors of 121?

1, 11, 121

There is a total of 3 positive divisors.
The sum of these divisors is 133.
The arithmetic mean is 44.333333333333.

3 odd divisors

1, 11, 121

How to compute the divisors of 121?

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 121 by each of the numbers from 1 to 121 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)

121 / 1 = 121 (the remainder is 0, so 1 is a divisor of 121)
121 / 2 = 60.5 (the remainder is 1, so 2 is not a divisor of 121)
121 / 3 = 40.333333333333 (the remainder is 1, so 3 is not a divisor of 121)
...
121 / 120 = 1.0083333333333 (the remainder is 1, so 120 is not a divisor of 121)
121 / 121 = 1 (the remainder is 0, so 121 is a divisor of 121)

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 121 (i.e. 11). 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 $\frac{N}{d} = D$ )

121 / 1 = 121 (the remainder is 0, so 1 and 121 are divisors of 121)
121 / 2 = 60.5 (the remainder is 1, so 2 is not a divisor of 121)
121 / 3 = 40.333333333333 (the remainder is 1, so 3 is not a divisor of 121)
...
121 / 10 = 12.1 (the remainder is 1, so 10 is not a divisor of 121)
121 / 11 = 11 (the remainder is 0, so 11 and 11 are divisors of 121)