Residues
 Phlembac Adib Hasan
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Re: Residues
Nice problem, Zan.Post more and more like these.
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 Residues.pdf
 I had to type it very swiftly.Please inform me if anybody can identify a typo.
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Re: Residues
Could you explain it, please? And did you mean $i\in N_n \backslash0$?Phlembac Adib Hasan wrote:So $f(i)\not\equiv0$(mod $n$) if $i\in N_n0 \ldots$
 Phlembac Adib Hasan
 Posts: 1016
 Joined: Tue Nov 22, 2011 7:49 pm
 Location: 127.0.0.1
 Contact:
Re: Residues
Yes.Also $N_{2^k}\equiv N_{2^{k+1}}N_{2^k}(mod\; 2^k)$ means $N_{2^k}\equiv N_{2^{k+1}}\backslash N_{2^k}(mod\; 2^k)$Zan wrote:Could you explain it, please? And did you mean $i\in N_n \backslash0$?Phlembac Adib Hasan wrote:So $f(i)\not\equiv0$(mod $n$) if $i\in N_n0 \ldots$
Like I said before, I was in a hurry and I avoided typing long codes like \backslash.
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Re: Residues
Thanks!
But could you explain me your argumentation, please?
I'm not very fast
But could you explain me your argumentation, please?
I'm not very fast
 Phlembac Adib Hasan
 Posts: 1016
 Joined: Tue Nov 22, 2011 7:49 pm
 Location: 127.0.0.1
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Re: Residues
Which part?
At first I showed we need to find those $n$s for which $S_n=\{\frac {0\cdot 1}{2},\frac {1.2}{2},...,\frac {(n1)n}{2}\}$ form a complete residue system modulo $n$.Hence there must be only one element in the set which is divisible by $n$.As we are seeing $n\frac {0\cdot 1}{2}$, so no other element in $S_n$ is divisible by $n$.Then I showed $n$ can't be odd because $n\frac {(n1)n}{2}$, another element of $S_n$ if $n$ is odd.Next I showed If $n=2^aq$($q$ odd>1), then we can find another element in $S_n$,different from $\frac {0\cdot 1}{2}$, which is divisible by $n$.(For example, if $n=12$ then $12\frac {8\times 9}{2}$.)At last I proved by induction that each positive integer power of $2$ satisfies the conditions.
At first I showed we need to find those $n$s for which $S_n=\{\frac {0\cdot 1}{2},\frac {1.2}{2},...,\frac {(n1)n}{2}\}$ form a complete residue system modulo $n$.Hence there must be only one element in the set which is divisible by $n$.As we are seeing $n\frac {0\cdot 1}{2}$, so no other element in $S_n$ is divisible by $n$.Then I showed $n$ can't be odd because $n\frac {(n1)n}{2}$, another element of $S_n$ if $n$ is odd.Next I showed If $n=2^aq$($q$ odd>1), then we can find another element in $S_n$,different from $\frac {0\cdot 1}{2}$, which is divisible by $n$.(For example, if $n=12$ then $12\frac {8\times 9}{2}$.)At last I proved by induction that each positive integer power of $2$ satisfies the conditions.
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 Phlembac Adib Hasan
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Re: Residues
I hate formal solution writing because it makes everything crazy. Notice that for each $i\in N_n$,\[i\equiv f(b_{n_{i}})(mod\; n).\]So $\{f\left (b_{n_{0}}\right ),f\left (b_{n_{1}}\right ),...,f\left (b_{n_{n1}}\right )\}$ makes complete residue set modulo $n$.Notice that all $b_{n_{i}}$s must be distinct modulo $n$ to follow the last condition.So for each $b_{n_{i}}$ there exists a unique $a_{n_{b_{n_{i}}}}\in N_n$ from the first definition.Hence it follows that \[f\left (b_{n_{i}}\right )\equiv f\left (a_{n_{b_{n_{i}}}}\right )(mod\; n)\].So it follows that $S_n$ is reduced residue system modulo $n$.Last part of $(i)$ is explained in my previous post.Let me know whehter it is clear enough now.
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Re: Residues
I want to give another solution:
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )