### 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?

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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?

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

Check this post for a brief introduction to latex: http://www.matholympiad.org.bd/forum/vi ... p?f=25&t=2

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 $\text{Modular Arithmatic}$.

https://docs.google.com/viewer?a=v&pid= ... MDZjYWNkMg

https://docs.google.com/viewer?a=v&pid= ... MDZjYWNkMg

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

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

Posted: **Wed Feb 21, 2018 9:53 pm**

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.

Posted: **Thu Mar 08, 2018 8:37 pm**

27 is correct answer.