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

This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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 :)

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?

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!!!!!

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

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

i got the last digit but not mod13 part

Log in to reply

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