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Tue 18 Oct, 2005 10:57 pm
Find the least positive residues of each of the following:
(a). 6! mod 7
(b). 10! mod 11
(c). 12! mod 13
(d). 16! mod 17
(e). Can you propose a theorem from the above congruences?
If you do (a) through (d), the answer to e is obvious. Note that all the values used for the modulus are prime. Non prime values yield zero except for when the modulus is 4.
Engineer
I agree on the conjecture on the prime modulus. But I don't think 4 is unique, special, maybe, as the 2nd even number, but not unique.
Rap
I think that four is unique as a non-prime that does not yield a zero modulus value. Here is my logic:
For non prime numbers that are not squares of a prime number, the factors making up that number are multiplied together in the factorial term, so (n-1)! mod n is zero.
When n is a perfect square of a prime number (k) then the factorial series doesn't contain k twice, but it does contain k and 2k. The product 2k^2 mod n=0, so those values also yield a zero modulus. The only case where a perfect square does not contain a multiple of k is where k is 2 and n=4.
OK. So what is the conjecture? I don't see it.
Does't this just lead to Wilson's Theorem:
(n - 1)! = -1 (mod n) ???
You know the answer, RK4.
satt_fs wrote:You know the answer, RK4.
Well, I wasn't sure. So it is Wilson's Thm? Right?
RK4 wrote:Does't this just lead to Wilson's Theorem:
(n - 1)! = -1 (mod n) ???
But only for prime values of n.