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    1. POfinding a/b mod c
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    1. USIan
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    1. CONo, B could never be large enough to make A/B into zero. It could be very large and A/B would be extremely small but it would never actually be zero. You can say it tends to zero as B tends to infinity though so your general idea was right. I'm just a mathematician so like to be precise. :) But that observation of taking a fixed A and varying B does admirably demonstrate the fact there are multiple valid answers. Its interesting to wonder if the correct answer is meant to be a function of m,n or if it just a badly specified question... +1 overall anyway. ;-)
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      1. USChris
    2. COEh? are you two using real numbers in an equation that specified "mod C"? In princple of course it's possible to specify a number system that works that way, but the normal convention is that you're working with integers modulo some integer, so no matter how large B is, it is also some value 0 <= n < C (the numbers represent infinite sets of integers), and while A/B will always have an integer result. Division is always defined as the inverse of multiplication, so where multiplication behaves differently (e.g. by giving a modulo C result), division behaves differently to match.
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      1. USSteve314
    3. COSpecific example: Mod-6 arithmetic. 3/2 is undefined. Why? Nothing you can multiply by two will yield an odd number, given the even modulus. Is that what you mean? Us computer guys are used to dealing with mod-2^32 arithmetic, except that division is defined as an approximation so that |b*(a/b)| <= |a|. Oh, and one other thing: He didn't say he was working in modular arithmetic; he said he had some numbers modulo another number. This is maybe splitting hairs, but should the presence of a modulus operation be enough to conclude that the entire problem is about modular arithmetic?
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