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

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!!!!!

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?

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

@Krishna Ar
–
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 :)

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

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i got the last digit but not mod13 part

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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!!!!!

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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?

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

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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

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

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