Does a/b mod m = (a mod m)/(b mod m)?
I am trying to find nCr mod m for very large numbers. If a/b mod m = (a mod m)/(b mod m) then think I will have solved my problem.
It is for Project Euler. I am using the nCr formula using factorials.
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nbrooks
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Andrew Koroluk
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If b = m, then you'll have a divide by zero. – poke Jan 02 '12 at 03:45
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3I don't think so. All you need is one counter-example to prove it wrong, so try a few different sets of numbers, like 19, 9, and 4. – Hot Licks Jan 02 '12 at 03:47
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Are `a` and `b` relatively prime with `m` ? – ypercubeᵀᴹ Jan 02 '12 at 20:54
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If this belongs anywhere, it is stackexchange maths – The Unfun Cat Oct 06 '12 at 20:46
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http://math.stackexchange.com/ – nbrooks Oct 07 '12 at 10:55
3 Answers
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No.
If you have a=8, b=2, m=2 then you have a/b mod m = 8/2 mod 2 = 4 mod 2 = 0
and (a mod m)/(b mod m) = (8 mod 2)/(2 mod 2) = 0/0 = NaN
NaN is not equal to 0.
Niet the Dark Absol
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This identity does not hold. Here is a counter-example:
Let a = 21, b = 7, m = 7.
Then (21/7) = 3 and 3 mod 7 = 3
Alternately, 21 mod 7 = 0 and 7 mod 7 = 0.
But 0 / 0 is undefined (and certainly not 3).
Thus your identity does not hold. However, I am almost certain that it will hold if m and b are relatively prime.
kaosjester
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1Indeed. If `b` and `m` are relatively prime, there is a `c` such that `b*c mod m = 1`. Then `(a/b) mod m == (a/b)*(c*b) mod m == (a*c) mod m == (a mod m) * c mod m` and similarly for `b mod m`. – Daniel Fischer Jan 02 '12 at 15:07
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You can use the following link to evaluate (a/b)mod m..... http://mathworld.wolfram.com/Congruence.html
The answer for evaluating is given at the end..
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1This is helpful but this should be a comment as it does not directly answer his question. – Austin Henley Oct 06 '12 at 20:45