## Rightmost two digits of 3^999

For students of class 11-12 (age 16+)
HABIBUL MURSALEEN
Posts: 2
Joined: Sun Apr 12, 2015 2:59 pm

### Rightmost two digits of 3^999

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

Posts: 1016
Joined: Tue Nov 22, 2011 7:49 pm
Location: 127.0.0.1
Contact:

### Re: Rightmost two digits of 3^999

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

tanmoy
Posts: 312
Joined: Fri Oct 18, 2013 11:56 pm

### Re: Rightmost two digits of 3^999

Take (mod $100$).You can read this to learn how to solve such kind of problems.The last part of the note is about $\text{Modular Arithmatic}$.
"Questions we can't answer are far better than answers we can't question"

abidul
Posts: 1
Joined: Wed Dec 27, 2017 10:36 pm

### Re: Rightmost two digits of 3^999

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 samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: Rightmost two digits of 3^999

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 It can easily be solved by using $\phi$ function.