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Old 2012-02-20, 07:30   #5
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"Mr. Tuch"
Dec 2007
Chennai, India

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I will rather certainly to be back within a while.

Originally Posted by CRGreathouse View Post
I'm not sure what you're saying here. 2^20 = 76 is not the identity element in Z/100Z. 2 divides 100, so no positive power of 2 can be the identity.
What I meant was that 76 is being the multiplicative identity element from among the 20 generated elements inside Z/100Z \equiv {4 (mod 100)} - {20 (mod 100)}

Originally Posted by CRGreathouse View Post
No, the identity is 3^5000 = 1.
The question was this, being stated as follows
While the group order for the element 2 (mod 10n) is being
given correctly to be
why not such a thing as this not hold out at all for the element 3 as well, as since?

Originally Posted by CRGreathouse View Post
3 has order 100 in Z/1000Z, so it can't generate more than 100 of the 1000 elements. It rather misses out 7, 11, 13, 17, 19, 21, 23, 29, ... as well as all multiples of 2 and 5.
Why is it being so? Thus, why does it not generate at all, all the elements for the forms 1, 3, 7, 9, 21, 23, 27, 29, 41, 43, 47, 49, 61, 63, 67, 69, 81, 83, 87, 89 (mod 200) at all, even hereby?
For this example, such that
1, 3, 9, 27, 41, 43, 49, 67, 81, 83, 89
which are being generated
7, 21, 23, 29, 47, 61, 63, 69, 87
they are not being generated at all,
first of all, at once, for this

Or otherwise that is there a somewhat providable possible explicit formula given in order to determine which elements it can be able to generate, rather?

For this example, it does not violate the following rule at all, if it is being possible to generate x (mod 200), then it is not being possible to generate (x+100) mod 200 at all, although rather this does not hold out, applicable for the values for (mod 10n) for the values for n > 4 at all.

The discrete logarithm from inside for the ring Z/(10n)Z can be done within the polynomial time algorithm itself, as follows:
for the element 2, so for 3, as well as again, since it cycles every 20 elements, for last 2 digits, they can be checked out for the 20 elements by making use for the brute force techniques for the first place itself, and then
thus for the subsequent decimal places, within 5 values for the same decimal digit itself, as follows

2m = x (mod 10n)
=> 2? = x+10nk (mod 10n+1), 0 \le k \le 9
? = m + r (10n/5*2n-2), 0 \le r \le 4

this is being the values for itself
this technique does not hold out, applicable for the prime order field in general at all.
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