Question

\[79^{79} \equiv N \pmod{100}\]\(N\) is a two digit number.

Answer 1

Fermat's Little Theorem?

Answer 2

\[79 \equiv 79 \equiv 0 \pmod{79} \]Which is kinda obvious.

Answer 3

And so we meet again.

Answer 4

Euler's Theorem says that \(\phi(79) = 79\) T_T

Answer 5

Yeah ;-)

Answer 6

Yeah? I was talking to the modular maths problem XD
LOL
Jk

Answer 7

So anyway, 79^5 = -1(mod 100)

Answer 8

How do you know again...

Answer 9

79^2 = 6241 = 41(mod 100)
79^3 = (41)(79)(mod 100) = 39(mod 100)
79^4 = (39)(79)(mod 100) = 81(mod 100)
79^5 = (81)(79)(mod 100) = 99(mod 100) = -1(mod 100)

Answer 10

So \(79^{10} \equiv 1 \pmod{100}\)?

Answer 11

Yeah, that's true....

Answer 12

Meaning 79^70 = 1(mod 100), too :D

Answer 13

Okay, oh.

Answer 14

So does that mean \(79^{80} \equiv 1 \pmod {100}\)

Answer 15

Ah, yeah.

Answer 16

Yes... But it's 79^79 you want, right?
79^75 = -1(mod 100)

Answer 17

\[79^{70} \times 79^{5} \times 79^{4} \pmod{100}\]AH!

Answer 18

mhmm...
79^70 is 1
79^5 is -1
79^4, refer to the process we did earlier...

Answer 19

\[1 \times -1 \times 81 \pmod{100}\]

Answer 20

Yeah, that seems right :D

Answer 21

\[19\pmod{100}\]

Answer 22

Bingo... :|

Answer 23

:-D