- #1
mrandersdk
- 246
- 1
Hello
Today I looked at something that seems like it should have a simple solution, but I may have looked at for too long, and can't solve it. My problem is as follows.
When you divide a whole number by a whole number n, then it is clear that the possible remainders are 0,1,2,...n-1. If you then look at 7 devided by 5 you get remainder 2, 14 by five gives 4, 21 by 5 gives 1, 28 by 5 gives 3, and 35 by 5 gives 0, then the pattern repeats. That is all possible remainders is achieved by the first five multiplums of seven. I know it must have something to do with 5 and 7 being coprime. Because 18 and 15 shows a different pattern.
In general is it possible to show that if m is lager than m and n and m are coprime, then the first n multiplums of m devided by n, will always achieve all the possible remainders?
Are there anyway to prove what will in general happen to the pattern if n and m are not Co prime. It seems to me that all remainders are achieved, but one need to remove the remainders, in which the number that devides n and m, devides.
Hope my two questions Makes sense.
Today I looked at something that seems like it should have a simple solution, but I may have looked at for too long, and can't solve it. My problem is as follows.
When you divide a whole number by a whole number n, then it is clear that the possible remainders are 0,1,2,...n-1. If you then look at 7 devided by 5 you get remainder 2, 14 by five gives 4, 21 by 5 gives 1, 28 by 5 gives 3, and 35 by 5 gives 0, then the pattern repeats. That is all possible remainders is achieved by the first five multiplums of seven. I know it must have something to do with 5 and 7 being coprime. Because 18 and 15 shows a different pattern.
In general is it possible to show that if m is lager than m and n and m are coprime, then the first n multiplums of m devided by n, will always achieve all the possible remainders?
Are there anyway to prove what will in general happen to the pattern if n and m are not Co prime. It seems to me that all remainders are achieved, but one need to remove the remainders, in which the number that devides n and m, devides.
Hope my two questions Makes sense.