ParthKohli 2 years ago $79^{79} \equiv N \pmod{100}$$$N$$ is a two digit number.

1. ParthKohli

Fermat's Little Theorem?

2. ParthKohli

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

3. terenzreignz

And so we meet again.

4. ParthKohli

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

5. ParthKohli

Yeah ;-)

6. terenzreignz

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

7. terenzreignz

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

8. ParthKohli

How do you know again...

9. terenzreignz

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)

10. ParthKohli

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

11. terenzreignz

Yeah, that's true....

12. terenzreignz

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

13. ParthKohli

Okay, oh.

14. ParthKohli

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

15. ParthKohli

Ah, yeah.

16. terenzreignz

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

17. ParthKohli

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

18. terenzreignz

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

19. ParthKohli

$1 \times -1 \times 81 \pmod{100}$

20. terenzreignz

Yeah, that seems right :D

21. ParthKohli

$19\pmod{100}$

22. terenzreignz

Bingo... :|

23. ParthKohli

:-D