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### Rightmost two digits of 3^999

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

Posted: 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

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

Posted: 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 $\text{Modular Arithmatic}$.

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

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

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