Loser66
  • Loser66
What is \(11^{644}(mod~~645)\) Please, help
Mathematics
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SOLVED
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chestercat
  • chestercat
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Loser66
  • Loser66
Yes, I am here. :)
ganeshie8
  • ganeshie8
645 = 3*5*43
ganeshie8
  • ganeshie8
Notice below : \(11^{644} \equiv (-1)^{644}\equiv 1 \pmod{3}\) \(11^{644} \equiv (1)^{644}\equiv 1 \pmod{5}\) \(11^{644} \equiv (11)^{42*15+14}\equiv 11^{14}\equiv 1 \pmod{43}\)

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ganeshie8
  • ganeshie8
that means \(3 \mid (11^{644}-1)\) \(5 \mid (11^{644}-1)\) \(43 \mid (11^{644}-1)\) since \(3, 5, 43\) are primes, it follows \(3*5*43 \mid (11^{644}-1)\) which is same as saying \[11^{644}\equiv 1\pmod{3*5*43}\]
Loser66
  • Loser66
Thank you so much. My question is how we know 11^(14) =1 mod 43 without calculator?
ganeshie8
  • ganeshie8
that is a good question... before that, did you get why \(11^{42*15} \equiv 1\pmod{43}\) ?
Loser66
  • Loser66
Yes
ganeshie8
  • ganeshie8
may i know why you think that reduces to 1 ?
Loser66
  • Loser66
because 43 is a prime, and (43,11) =1 hence \(11^{42}\equiv 1(mod 43)\) and that leads to \(11^ {42*15}\equiv 1(mod 43)\)
Loser66
  • Loser66
Fermat told me that :)
Zarkon
  • Zarkon
you shouldn't talk to the dead
Loser66
  • Loser66
He "told" me through the book. :)
ganeshie8
  • ganeshie8
lol nice, do you feel \(11^{14}\) is hard to reduce ?
Loser66
  • Loser66
Yes.
ganeshie8
  • ganeshie8
how ?
ganeshie8
  • ganeshie8
show me how it is hard
ganeshie8
  • ganeshie8
because, i have figured out the answer in 2-3 steps working it in my head...
Loser66
  • Loser66
11^14 = 11*11*11*........(14 times) . I don't know how to make it easier.
ganeshie8
  • ganeshie8
Alright, here it is : \[11^{14} \equiv 121^7 \equiv (-8)^7\equiv - 2^{3*7}\equiv -128^3 \equiv -(-1)^3 \equiv 1\pmod{43}\]
Loser66
  • Loser66
oh, it is much easier. Thank you so much. I got it.

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