# Last digit problem

how to solve this please give a solution for it i have found it on internet $${ 7 }^{ { 7 }^{ { 7 }^{ 7... } } }(2008\quad 7's)mod(13)$$ and also its last digit

Note by Rishabh Jain
4 years, 1 month ago

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oh god !! so much discussion but no proper solution ??

- 4 years, 1 month ago

i got the last digit but not mod13 part

- 4 years, 1 month ago

Hi! Well, you can do this easily too! Come-on! Write the rem's obtained when 7 power 1-12 are divided by 13. You will find the numbers 1-12 are cycled. Simply plug-in the value of rem$$7^{7}$$ from there. That's all. Is 6 correct? Tell me where you got this from also! Please!!!!!

- 4 years, 1 month ago

Hey! You should see my comment below! It explains everything. If you still dont get it, please tell me...I will post the whole solution. And ,could you tell me the source of this problem pleasE?

- 4 years, 1 month ago

hey @Finn Hulse @Sharky Kesa @Krishna Ar or anyone please answer it !!

- 4 years, 1 month ago

Whoa! Thanks for tagging me here @Rishabh Jain . Well the answer is that the expression is 6 mod 13 and also 3 mod 10 (last digit). Tell me if its right, if yes, I shall very soon post a full solution. I can perhaps give a small hint that you must analyse the remainders obtained on divison of powers of 7 by 13 till u get a pattern/chain. Or plug in euler :p

- 4 years, 1 month ago

Ur answer is correct krishna. ...All hail euler :D

- 4 years, 1 month ago

:D HOW R YOU SO SURE? . I didnt use totient though

- 4 years, 1 month ago

I used totient to find last digit ..but still I am confused on how to deal with the 7^7 2008 times mod 13 part ...can u please give a hint :)

- 4 years, 1 month ago

Hint is above in the previous comment.

- 4 years, 1 month ago

Got it....but don't u think that it's a long process ....isn't there a shorter way ?

- 4 years, 1 month ago

Know wut? It is not long at all, I did it in some 45 secs! Srsly! No kiddin'

- 4 years, 1 month ago

Did u use eulers criterion too ? I tried using it ...I also got it in 45 sec

- 4 years, 1 month ago

No

- 4 years, 1 month ago

Well..y don't u post ur solution :/

- 4 years, 1 month ago

Gosh! Please see the very first comment of mine!

- 4 years, 1 month ago