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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##?
This notation is a bit odd, as it implies that n belongs to the real interval. A better way to write it IMO would beLet n∈[0,36)
I would probably do this first, rather than as an alternate approach.Alternatively, you can look at what n should be such that 48*n contains the factors of 36 namely 2*2*3*3.