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

 Jul 7, 2021

Best Answer 

 #1
avatar+2122 
+2

If we add all the exponents up, we get 3^(1+2+3+4... 43) = 3^(946). 

 

3, 9, 27, 81, 243, 729...

3, 9, 5, 4, 1, 3....

There is a cycle of 3, 9, 5, 4, 1 in mod 11. 

946 = 1 mod 11, and the first number in the cycle of 3, 9, 5, 4, 1 is 3. 

 

=^._.^=

 Jul 7, 2021
 #1
avatar+2122 
+2
Best Answer

If we add all the exponents up, we get 3^(1+2+3+4... 43) = 3^(946). 

 

3, 9, 27, 81, 243, 729...

3, 9, 5, 4, 1, 3....

There is a cycle of 3, 9, 5, 4, 1 in mod 11. 

946 = 1 mod 11, and the first number in the cycle of 3, 9, 5, 4, 1 is 3. 

 

=^._.^=

catmg Jul 7, 2021
 #2
avatar+26139 
+2

Find the remainder when \(3*3^2*3^3*\ldots*3^{43}\) is divided by \(11\).

 

\(\begin{array}{|rcll|} \hline 3*3^2*3^3*\ldots*3^{43} &=& 3^{1+2+3+\ldots 42+43} \\ &=& \left.3\right.^{\left(\frac{1+43}{2}*43\right)} \\ &=& \left.3\right.^{\left(\frac{44}{2}*43\right)} \\ &=& 3^{22*43} \\ &=& 3^{946} \\ \hline \end{array}\)

 

\(\begin{array}{lrcll} \text{Euler:}\\ \boxed{3^{\phi(11)}\equiv 1 \pmod{11} \quad | \quad \phi(11)=10,~ \phi(p) = p-1\\3^{10}=1 \pmod{11} } \\ \end{array}\)

 

\(\begin{array}{|rcll|} \hline 3^{946} \pmod{11} &\equiv& 3^{10*94+6}\pmod{11} \\ &\equiv& \left(3^{10} \right)^{94}3^6\pmod{11} \quad | \quad 3^{10} \equiv 1\pmod{11} \\ &\equiv& 1^{94}3^6\pmod{11} \\ &\equiv& 3^6\pmod{11} \\ &\equiv& 729\pmod{11} \\ \mathbf{3^{946} \pmod{11}} &\equiv& \mathbf{ {\color{red}3}\pmod{11}} \\ \hline \end{array}\)

 

laugh

 Jul 8, 2021

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