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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 beMr Davis 97 said:Let n∈[0,36)
I would probably do this first, rather than as an alternate approach.jedishrfu said:Alternatively, you can look at what n should be such that 48*n contains the factors of 36 namely 2*2*3*3.
To enumerate integers n s.t. 36 | 48n means to list out all possible values of n that satisfy the condition that 36 is a multiple of 48n. In other words, 36 is evenly divisible by 48n without any remainder.
The number 36 is significant because it is the constant, or divisor, in the given condition. In order for 36 to be a multiple of 48n, n must be a factor of 36.
Some examples of integers n that satisfy the condition 36 | 48n are 1, 2, 3, 4, 6, 9, 12, 18, and 36. These are all factors of 36, and when multiplied by 48, result in a multiple of 36.
Yes, there are restrictions. In order for 36 | 48n to be true, n must be a positive integer. It cannot be a negative number or a fraction. Additionally, n cannot be 0 since any number multiplied by 0 is 0, which would not satisfy the condition.
There are infinitely many integers n that satisfy the condition 36 | 48n. This is because there are an infinite number of factors of 36, and each of these factors can be multiplied by 48 to result in a multiple of 36. Therefore, the list of integers n will continue indefinitely.