Enumerating integers n s.t. 36 | 48n

[Mr Davis 97]

This is a simple computational question. Let $n \in [0, 36)$. What's the fastest way to list all $n$ s.t $36$ divides $48n$?

 

[jedishrfu]

Well first I would get a list of numbers for n where it is true and from there deduce the fastest algorithm to list them.

Alternatively, you can look at what n should be such that 48*n contains the factors of 36 namely 2*2*3*3.

so what must n provide so that the product contains the factors of 36?

 

[Mark44]

Let n∈[0,36)

This notation is a bit odd, as it implies that n belongs to the real interval. A better way to write it IMO would be
$\text{Let } n \in \{0, 1, 2, \dots, 35\}$

Alternatively, you can look at what n should be such that 48*n contains the factors of 36 namely 2*2*3*3.

I would probably do this first, rather than as an alternate approach.