Rightmost two digits of 3^999

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HABIBUL MURSALEEN
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Rightmost two digits of 3^999

Unread post by HABIBUL MURSALEEN » Thu Dec 03, 2015 8:06 am

What are the rightmost two digits of $3^{999}$ ? What's the process in problems of this kind? :roll:
Last edited by Phlembac Adib Hasan on Mon Dec 07, 2015 10:56 pm, edited 1 time in total.
Reason: Latexed

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Phlembac Adib Hasan
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Re: Rightmost two digits of 3^999

Unread post by Phlembac Adib Hasan » Mon Dec 07, 2015 11:11 pm

Hello, welcome to this forum. Please use latex for maths in future. It honestly makes our work a lot easier.
Check this post for a brief introduction to latex: http://www.matholympiad.org.bd/forum/vi ... p?f=25&t=2
Welcome to BdMO Online Forum. Check out Forum Guides & Rules: http://forum.matholympiad.org.bd/viewtopic.php?f=25&t=6

tanmoy
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Re: Rightmost two digits of 3^999

Unread post by tanmoy » Mon Dec 07, 2015 11:40 pm

Take (mod $100$).You can read this to learn how to solve such kind of problems.The last part of the note is about $\text{Modular Arithmatic}$.
https://docs.google.com/viewer?a=v&pid= ... MDZjYWNkMg
I like girls and mathematics; both are beautiful.

abidul
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Re: Rightmost two digits of 3^999

Unread post by abidul » Thu Jan 04, 2018 12:59 am

firstly, we can get 3^15=== 7 (mod100) =>3^990 ===7^66
Now,7^4=== 1(mod 100) =>7^66 === 1*7^2 = 49
and, 3^9=== 83 (mod100)
So, 3^999 = 3^990 *3^9=== 49*83 = 4067 === 67
Rightmost two digits of 3^999 is 7 and 6 :oops:

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samiul_samin
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Re: Rightmost two digits of 3^999

Unread post by samiul_samin » Wed Feb 21, 2018 9:53 pm

abidul wrote:
Thu Jan 04, 2018 12:59 am
firstly, we can get 3^15=== 7 (mod100) =>3^990 ===7^66
Now,7^4=== 1(mod 100) =>7^66 === 1*7^2 = 49
and, 3^9=== 83 (mod100)
So, 3^999 = 3^990 *3^9=== 49*83 = 4067 === 67
Rightmost two digits of 3^999 is 7 and 6 :oops:
Please,use LATEX. It makes your solution more readable.
This is BdMO National Higher Secondary 2006 problem.
It can easily be solved by using $\phi$ function.
$\int^{\infty}_0 e^{-x} dx=1$

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