## Rightmost two digits of 3^999

For students of class 11-12 (age 16+)
HABIBUL MURSALEEN
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### 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

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### 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
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tanmoy
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### 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
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### 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
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### 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
This is BdMO National Higher Secondary 2006 problem.
It can easily be solved by using $\phi$ function.

Mathlomaniac
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