mindmind wrote:
Please help me understand better :
The remainder of 4^40 would be same as that of 4 .
Is my approach wrong? Can you provide similar examples, where it is not.
In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)).
\(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\).
Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\).
\(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\).[/quote]
Yes, Agreed
So I should consider : Something near to Remainder 1
Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime
eg : 4^40 divided by 7 give a remainder of 4.[/quote]
I am not sure what you mean here:
Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime
eg : 4^40 divided by 7 give a remainder of 4.\(3^4=81=11\cdot{7}+4\). \(7\) is prime, but the remainder is neither \(3\), nor a prime.
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