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# PLZ HELP NOw

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What is the residue modulo 13 of the sum of the modulo 13 inverses of the first 12 positive integers?

Express your answer as an integer from 0 to 12 , inclusive.

Dec 9, 2018

#1
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I typed this in calc: $$tan(90)$$

and it was equal to $$infinity$$.

Dec 9, 2018
#2
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You need to post a new question of your own.

you have asked a good question ... but you need to post it properly.

Dec 9, 2018
#3
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What is the residue modulo 13 of the sum of the modulo 13 inverses of the first 12 positive integers?
Express your answer as an integer from 0 to 12 , inclusive.

$$\begin{array}{|rcll|} \hline 1\cdot { \color{red}1} &\equiv& 1 \pmod{13} \\ 2\cdot 2^{-1} \equiv \pmod{13} \equiv 2\cdot { \color{red}2^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 3\cdot 3^{-1} \equiv \pmod{13} \equiv 3\cdot { \color{red}3^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 4\cdot 4^{-1} \equiv \pmod{13} \equiv 4\cdot { \color{red}4^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 5\cdot 5^{-1} \equiv \pmod{13} \equiv 5\cdot { \color{red}5^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 6\cdot 6^{-1} \equiv \pmod{13} \equiv 6\cdot { \color{red}6^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 7\cdot 7^{-1} \equiv \pmod{13} \equiv 7\cdot { \color{red}7^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 8\cdot 8^{-1} \equiv \pmod{13} \equiv 8\cdot { \color{red}8^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 9\cdot 9^{-1} \equiv \pmod{13} \equiv 9\cdot { \color{red}9^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 10\cdot 10^{-1} \equiv \pmod{13} \equiv 10\cdot { \color{red}10^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 11\cdot 11^{-1} \equiv \pmod{13} \equiv 11\cdot { \color{red}11^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 12\cdot 12^{-1} \equiv \pmod{13} \equiv 12\cdot { \color{red}12^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ \hline \end{array}$$

$$\begin{array}{|ll|} \hline & \text{sum}_{\text{of the modulo 13 inverses of the first 12 positive integers}} \\ =& 1^{11}+2^{11}+3^{11}+4^{11}+5^{11}+6^{11}+7^{11}+8^{11}+9^{11}+10^{11}+11^{11}+12^{11} \pmod{13} \\ =& 1+7+9+10+8+11+2+5+3+4+6+12 \pmod{13} \\ =& 1+2+3+4+5+6+7+8+9+10+11+12 \pmod{13} \\ =& \dfrac{1+12}{2}\cdot 12 \pmod{13} \\ =& 13 \cdot 6 \pmod{13} \\ \mathbf{=}& \mathbf{0 \pmod{13}} \\ \hline \end{array}$$

Dec 10, 2018
edited by heureka  Dec 10, 2018