When is m33 divisible by 36? (m is an integer).

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To determine when m33 is divisible by 36, it is essential to find the order of the element 33 in Z_{36}. The equation 33n = 36m can be rearranged to n = 36m/33, indicating that n must be an integer for divisibility. The modular equivalence shows that 33 is congruent to -3 mod 36, leading to the conclusion that the order of 33 is 12. Similarly, for the element 15, the analysis reveals that its order is also 12, as both elements share the same factor requirements for divisibility by 36. Understanding the prime factorization of 36 and how it relates to the elements in Z_{36} is crucial for solving these types of problems.
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Homework Statement



Work out the order of the following elements;

33 \in Z_{36}

The Attempt at a Solution



It's probably really simple. But this only happens when an integer times 33 is divisible by 36.

That is;

33n = 36m

Which I can re-arrange to find

n = 36m/33

Now, I can keep adding 36/33 in my calculator until I get an integer result, but surely there is an easier way? For example;

33 \equiv -3mod36

Which suggests that 36/3 = 12 is the order of 33. This is fine and dandy, but what happens when I get to a question like, find the order of

15 \in Z_{36}

Then

15 \equiv -21mod36

Which means I'm back to the same problem again. My modular arithmetic is fairly poor, so how would I work this out?
 
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Factor 36 into prime factors and think about what factors k must have for 36 to divide 33k.
 
36 = 3x3x2x2

33 = 11x3

So k must be 3x2x2 = 12.

Also

15 = 5x3

So k must be 3x2x2 = 12.


Thanks!
 
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