anonymous
  • anonymous
How do I determine 8^402 mod 5? I know that 8 is 3 mod 5 but I get stuck after that.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
i believe 5 and 402 are relatively prime
amistre64
  • amistre64
8*8*8*...*8 = x (m5) 3*3*3*...*3 = x (m5) 402 times 9*9*9*...*9 = x (m5) 201 times 4*4*4*...*4 = x (m5) 201 times 16*16*16*...*16 * 4 100 times 1 time 16 = 1 (m5) so im thinking we get down to 4 (m5) in the end
amistre64
  • amistre64
theres prolly some thrms that ive forgotten about that could make that simpler overall

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amistre64
  • amistre64
\[8^{402}\] \[4^{402}~2^{402}\] \[(4^2)^{201}~2^{402}\] \[(16)^{201}~2^{402}\] \[(1)^{201}~2^{402}\] \[2^{402}\] \[2^{400+2}\] \[2^{400}~2^2\]
anonymous
  • anonymous
the answers i'm finding say 4 mod 5 but I just don't understand how it gets there
amistre64
  • amistre64
it has to do with reducing the setup ... if you want to approach it the long way
amistre64
  • amistre64
there is a shorter route that i can never fully recall; but i think that 2^(400) 2^(2) is key to it
amistre64
  • amistre64
my long route, was to take this to 4^2 = 16; since 16 = 5(3)+1 , 16 = 1 mod5 1*1*1*1*1 ... = 1 so all that leaves us the the *4 thats left over 4 = 4 mod5
amistre64
  • amistre64
8*8 = 64 = -1 mod 5 would have been another way to view it \[8^{402}\] \[(8^2)^{201}\] \[(64)^{201}\] \[(-1)^{201}\] \[(-1)^{200+1}\] \[(-1)^{200}~(-1)^1\] \[(-1)^1=-1\] -1 = 5(-1) + 4, therefore ... 4 mod 5

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