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Find the remainder when 3*3^2*3^3*...*3^43*3^44*3^45 is divided by 11.

 Feb 6, 2022
 #1
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productfor(n, 1, 45, 3^n)  mod 11==1 - the remainder.

 Feb 6, 2022
 #2
avatar+117861 
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Thanks guest,

 

Here is a neat way to do it.

 

\(3*3^2*3^3*...*3^{43}*3^{44}*3^{45}\mod(11)\\ \text{Lets look for a pattern}\\ 3^1=3\\ 3^2=9\mod(11)=-2\mod(11)\\ 3^3=-2*3\mod(11)=-6\mod(11)=5\mod(11)\\ 3^4=5*3\mod(11)=4\mod(11)\\ 3^5=4*3\mod(11)=1\mod(11)\\ 3^6=1*3\mod(11)=3\mod(11)\\ \text{The pattern has started to repeat.} \)

there is 45 terms, 5*9 =45 so I will have to rasie the procuct to a power of 9

 

so I have  

(3*-2*5*4*1) ^9    mod(11)

=(-120)^9   mod(11)

=(-121+1)^9  mod(11)

=(1)^9  mod(11)

 

=1   mod(11)

 

 

 

 

 

 

LaTex

3*3^2*3^3*...*3^{43}*3^{44}*3^{45}\mod(11)\\
\text{Lets look for a pattern}\\
3^1=3\\
3^2=9\mod(11)=-2\mod(11)\\
3^3=-2*3\mod(11)=-6\mod(11)=5\mod(11)\\
3^4=5*3\mod(11)=4\mod(11)\\
3^5=4*3\mod(11)=1\mod(11)\\
3^6=1*3\mod(11)=3\mod(11)\\
\text{The pattern has started to repeat.}
 

 Feb 6, 2022

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