Question

How would I find the last two digits of 79^79 ?
I know that 79^79 mod 100 = 79^39 mod 100 since:
79 mod phi(100) = 79 mod 40 = 39. But after that I'm lost. Help would be appreciated.

Answer 1

Brilliant.org eh ? ;)

Answer 2

Yes ;)
It was a bad example sorry. I've read the blog but I still don't get it.

Take something else. Last two digits of 13^13 ?

Answer 3

I'll hint you
79^5 = -1 mod100

Answer 4

Well I am not an expert at modular arithmetic, but my approach for this one was to find a power of 79 close to 100, manually, 79^5 was one

Answer 5

Sounds reasonable, I'll try it. Thanks

Answer 6

Well, well, something looks eerily familiar :D
How'd you know 79^5 was -1(mod 100), @shubhamsrg
XD

Answer 7

As I said, I manually calculated it, used a calculator

Answer 8

sounds masochistic :D

Answer 9

hmm, in case you have a shorter way ?

Answer 10

Reminds me of this one
http://openstudy.com/study#/updates/510a825ae4b070d859bf144e

Answer 11

No, I was just reminiscing... good times XD

Answer 12

Okey the n^k = -1 mod 100 seems to be important but why and how?

Answer 13

hmm, maybe I see something

79 = 80-1

79^79 = (80-1)^79

All terms on expansion, of RHS will be divisible by 100, except ,

C(79,0) * (-1) + C(79,1) * (80)

= (79*80)-1

omg it was so easy! :O

Answer 14

Not important, maybe lucky :D
Because if
a = b mod 100
a^n = b^n mod 100

And I don't know about you, but raising -1 (or 1) for that matter to exponents seems like a relatively simple task :D

Answer 15

aha now i see.
79^5^8 = -1^8 mod 100
79^40 = 1 mod 100
cool

Answer 16

I wasted a lot of time on this ques when I was myself attempting it a few hours before.

Now it seems a lot easier

79= 80-1
Thats it.