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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestoh god !! so much discussion but no proper solution ?? – Rishabh Jain · 2 years, 5 months ago

Log in to reply

– Rishabh Jain · 2 years, 5 months ago

i got the last digit but not mod13 partLog in to reply

– Krishna Ar · 2 years, 5 months 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!!!!!Log in to reply

– Krishna Ar · 2 years, 5 months 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?Log in to reply

hey @Finn Hulse @Sharky Kesa @Krishna Ar or anyone please answer it !! – Rishabh Jain · 2 years, 5 months ago

Log in to reply

@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 – Krishna Ar · 2 years, 5 months ago

Whoa! Thanks for tagging me hereLog in to reply

– Poonayu Sharma · 2 years, 5 months ago

Ur answer is correct krishna. ...All hail euler :DLog in to reply

– Krishna Ar · 2 years, 5 months ago

:D HOW R YOU SO SURE? . I didnt use totient thoughLog in to reply

– Poonayu Sharma · 2 years, 5 months 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 :)Log in to reply

– Krishna Ar · 2 years, 5 months ago

Hint is above in the previous comment.Log in to reply

– Poonayu Sharma · 2 years, 5 months ago

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

– Krishna Ar · 2 years, 5 months ago

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

– Poonayu Sharma · 2 years, 5 months ago

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

– Krishna Ar · 2 years, 5 months ago

NoLog in to reply

– Poonayu Sharma · 2 years, 5 months ago

Well..y don't u post ur solution :/Log in to reply

– Krishna Ar · 2 years, 5 months ago

Gosh! Please see the very first comment of mine!Log in to reply