## ParthKohli 2 years ago A common technique for finding modulo $$10$$ of any huge number in the form $$a^b$$?

1. ParthKohli

It seems that you have to go through a lot of manipulations... a lot for doing that. Although I was thinking of a technique they use to figure all that out.

2. ParthKohli

OK, I'm gonna change the topic to something else:$39^{39} \equiv n\pmod{100}$

3. ParthKohli

$$n$$ has to be a two-digit number. (I actually have to find the last two digits)

4. ParthKohli

Something with a pattern?

5. shubhamsrg

39 = -61 mod100 =>39^38 = 61^38 mod100 =>39^39 = 39*61^38 mod100 SInce 61^n always ends in 1, that multiplied by 39, will give last digit as 9, hmm, leme see for the second last.

6. terenzreignz

Better yet, 39^5 = -1 mod 100 Use that: 39^39 = (39^5)^7 * 39^4 = (-1)^7 * 39^4 mod 100

7. shubhamsrg

Seems legit. But on paper, who's gonna go till 39^5 ! :O

8. ParthKohli

Very nice, can you tell me *how* you came up with that solution? What do you initially think?

9. terenzreignz

square 39, and then only take the last two digits...

10. terenzreignz

then multiply 39 to it again.

11. terenzreignz

and only take the last two digits, again...

12. ParthKohli

Ahem, that was beautiful. But is there a common way to figure out these solutions?

13. shubhamsrg

Ohh, I see.

14. terenzreignz

The way I think of things, and it's a common trend that the way I do it is highly inefficient (LOL) But when faced with a problem b^p = n (mod k) I try to see if there's an exponent to which b can be raised so that it is 1 (or -1) mod k first.

15. terenzreignz

That "last two digits" trick only works because you were conveniently working under mod 100... XD

16. ParthKohli

Nice... can you give me another example?

17. terenzreignz

How about 6^5000 = n (mod 215)

18. terenzreignz

(sneakily thinking of simple problems) :>

19. ParthKohli

Arhmegurd.$\left(6^{5}\right)^{1000} \equiv (6^2)^{1000} \equiv n \pmod{215}$

20. ParthKohli

$\left(6^5\right)^{400} \equiv (6^2)^{400} \equiv n\pmod{215}$

21. ParthKohli

$\left(6^5\right)^{120} \equiv 6^{240} \equiv n \pmod{215}$

22. terenzreignz

I'm not following o.O You may be using a method I'm not familiar with, though... if you are, do continue :)

23. ParthKohli

$\left(6^{5}\right)^{48} \equiv 6^{96} \equiv n \pmod{215}$

24. ParthKohli

Hmm... I am just using $$6^2 \equiv 6^5 \pmod{215}$$

25. ParthKohli

What should I do here?

26. terenzreignz

Hang on...

27. terenzreignz

I was rather hoping you'd go for $6^3 = 216 = -1(mod \ 215)$

28. ParthKohli

Oh...

29. terenzreignz

Wait, typo -.-

30. ParthKohli

Yes, isn't that 1 (mod 215)

31. terenzreignz

$\large 6^3 = 216 = 1(\mod \ 215 \ \ \ \ )$ Much better

32. terenzreignz

And work from there...

33. terenzreignz

$\large 6^{5000} = 6^{3(1666)+2}$

34. ParthKohli

$6^{5000}= 6^{4998} \times 36 \equiv$

35. terenzreignz

$1 \times 36 (\mod 215 \ \ \ \ \ )$

36. ParthKohli

$\equiv 36 \pmod{215}$

37. ParthKohli

Argh, you type fast. =3

38. terenzreignz

Awesomeness

39. ParthKohli

Another n00bish question coming up. Thanks man.

40. terenzreignz

If you have time, look up Fermat's Little Theorem, might come in handy in a pinch.

41. ParthKohli

Yesh, that one is cool:$p^{n - 1} \equiv 1 \pmod{n}$I'd try to see if that applies in the future.