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## Main Question or Discussion Point

I'll call it the "Wheel Lug Lemma" for now.

If there are a pair of integers

The idea is if there is a wheel with

For example, typically a wheel has q = 5 lug nuts, and it is recommended that they be tightened in a star pattern, so that they are done in the order of 0, 2, 4, 1, 3, and thus with p = 2 skipping.

0 % 5 = 0

2 % 5 = 2

4 % 5 = 4

6 % 5 = 1

8 % 5 = 3 → all possible remainder have been cycled through

10 % 5 = 0 → the cycle repeats

I figure that someone must have recognized this and wrote it up as a lemma somewhere.

If there are a pair of integers

*p*&*q*such that the Greatest Common Denominator is 1, and there is some number*s*that is product of*p*and an increasing whole number*n*, then the remainder of the division of*s*by*q*will cycle through all values of from*0*up to, but not including*q*until*n*is equal to*q*at which time the remainder is*0*, with the cycle repeating again in the same order.The idea is if there is a wheel with

*p*# of lug nuts, and the lug nuts are tightened in the order skipping*q*# of lug nuts to tighten the next nut, then eventually every lug nut will get tightened before encountering one that has already been tightened.For example, typically a wheel has q = 5 lug nuts, and it is recommended that they be tightened in a star pattern, so that they are done in the order of 0, 2, 4, 1, 3, and thus with p = 2 skipping.

0 % 5 = 0

2 % 5 = 2

4 % 5 = 4

6 % 5 = 1

8 % 5 = 3 → all possible remainder have been cycled through

10 % 5 = 0 → the cycle repeats

I figure that someone must have recognized this and wrote it up as a lemma somewhere.