Friday, 29 April 2016

elementary number theory - Is this proof for 1110equiv1pmod100 correct?



Show that 11101(mod100)



here, I asked a question about this same problem and all the answers say that I prove this by binomial expansion.



But by doing binomial expansion to show this is the same as doing 1110 mod  100 in the calculator, no property is being showed to prove this.



But I think this proof might be right.




gcd(11,100)=1



11^{φ(100)} \equiv 1 \pmod{100}



11^{40} \equiv 1 \pmod{100} \ ⇒



11^{10} *11^{10} *11^{10}* 11^{10}\equiv 1\pmod{100}



by Looking everybody mod \ 100, I can say that 11^{10} \equiv 1\pmod{100}



Answer



This doesn't work because the solutions to x^4 \equiv 1 \pmod{100} are x \equiv 1, 7, 43, 49, 51, 57, 93, 99 \pmod{100}. So from (11^{10})^4 \equiv 1 \pmod{100} the best you can say is that 11^{10} \equiv 1, 7, 43, 49, 51, 57, 93, \text{ or } 99 \pmod{100}.


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...