20120219, 17:02  #1 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
2/3 Powers being viewed over the Ring Z/(10^n)Z
I am doing very severe load coursework for Ph.D. from IMSc, Chennai,
that's why I am not able tending to be active over this forum for the past year at all. Very recently, I had come across some very fascinating property, for this, I'd like (need) to seek the answer for this. Euler's Theorem conveys necessarily that the order for an element over (mod n) is always being a divisor for , but though the properties for the powers of 2 & 3 vary accordingly as follows. Why is it being so, the properties are rather being different from them apart? As follows => is being the identity element over Z/100Z => is being the identity element over Z/1000Z ... Consider this, rather => is Not being the identity element at all => WHY? 
20120219, 17:04  #2 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
Next, for the powers of 2, over the ring Z/1000Z, all the 100 elements for the form 8 (mod 1000), 40 (mod 1000) are being generated.
Unlike this, for the powers of 3, although over the ring Z/100Z all the 20 elements 1, 3, 7, 9 (mod 20) are being generated, this does not hold out over on from the insider for the larger ring Z/1000Z at all. For this example, the element 3, over on multiplication yields, generating the elements 1, 3, 9 (mod 1000), although the element 7 (mod 1000) is not being generated at all, although, instead it rather generates the element 507 (mod 1000), although, rather. AGAIN WHY? What is being it to be the true reason behind this, rather? WHY? Last fiddled with by Raman on 20120219 at 17:24 
20120220, 02:42  #3 
Aug 2006
3×5^{2}×79 Posts 
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.
No, the identity is 3^5000 = 1. 
20120220, 02:47  #4  
Aug 2006
3·5^{2}·79 Posts 
Quote:


20120220, 07:30  #5  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3·419 Posts 
I will rather certainly to be back within a while.
Quote:
The question was this, being stated as follows While the group order for the element 2 (mod 10^{n}) is being given correctly to be 10^{n}/5*2^{n2}, why not such a thing as this not hold out at all for the element 3 as well, as since? Quote:
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 10^{n}) for the values for n > 4 at all. The discrete logarithm from inside for the ring Z/(10^{n})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 2^{m} = x (mod 10^{n}) => 2^{?} = x+10^{n}k (mod 10^{n+1}), 0 k 9 ? = m + r (10^{n}/5*2^{n2}), 0 r 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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