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There are four positive integers a,b,c,d less than 8 which are invertible modulo 8. Find the remainder when (abc+abd+acd+bcd)(abcd)^{-1} is divided by 8.

Guest Jul 29, 2018
 #1
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First, we find the four numbers, which are 1, 3, 5 and 7. We then expand (abc+abd+acd+bcd)(abcd)^(-1) to get a^(-1)+b^(-1)+c^(-1)+d^(-1). Plugging in 1, 3, 5, and 7, we get 1^(-1)+3^(-1)+5^(-1)+7^(-1) which is congruent to 1+3+5+7 mod 8. Adding them together, we see that 1+3+5+7=16, which is congruent to 0 mod 8.\( \)

Guest Jul 29, 2018
 #2
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There are four positive integers a,b,c,d less than 8 which are invertible modulo 8.

Find the remainder when (abc+abd+acd+bcd)(abcd)^{-1} is divided by 8.

\(\begin{array}{|rcll|} \hline && (abc+abd+acd+bcd)(abcd)^{-1} \\ &=& \dfrac{abc}{abcd} +\dfrac{abd}{abcd}+\dfrac{acd}{abcd}+\dfrac{bcd}{abcd} \\ &=& \dfrac{1}{d} +\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a} \\ &=& a^{-1}+b^{-1}+c^{-1}+d^{-1} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcl|} \hline \gcd(1,8)=\gcd(3,8)=\gcd(5,8)=\gcd(7,8)=1 \quad & | \quad a=1\quad b=3\quad c=3\quad d=5 \\\\ \begin{array}{|rcl|} \hline \phi(8)&=& 8\cdot\left(1-\dfrac12 \right) \\ &=& 4 \\ \hline \end{array} \\ \hline \end{array}\)

 

\(\small{ \begin{array}{|rcll|} \hline && \Big( 1^{-1} \pmod {8} + 3^{-1} \pmod {8} + 5^{-1} \pmod {8} + 7^{-1}\pmod {8} \Big)\pmod {8} \\\\ &=& \Big( 1^{\phi(8)-1} \pmod {8}+ 3^{\phi(8)-1} \pmod {8} + 5^{\phi(8)-1} \pmod {8} + 7^{\phi(8)-1}\pmod {8} \Big) \pmod {8} \\\\ &=& \Big( 1^{4-1} \pmod {8}+ 3^{4-1} \pmod {8} + 5^{4-1} \pmod {8} + 7^{4-1}\pmod {8} \Big) \pmod {8} \\\\ &=& \Big( 1^{3} \pmod {8}+ 3^{3} \pmod {8} + 5^{3} \pmod {8} + 7^{3}\pmod {8} \Big) \pmod {8} \\\\ &=& ( 1^{3}+ 3^{3} + 5^{3} + 7^{3} ) \pmod {8} \\\\ &=& ( 1+ 27 + 125 + 343 ) \pmod {8} \\\\ &=& 496 \pmod {8} \\\\ &=& 62\cdot 8 \pmod {8} \\\\ &\mathbf{=}&\mathbf{ 0 \pmod {8} } \\ \hline \end{array} }\)

 

laugh

heureka  Jul 30, 2018

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