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Kazeer [188]
3 years ago
8

Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$

is divided by 7?
Mathematics
1 answer:
Blizzard [7]3 years ago
6 0
\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}

a^{81}\equiv1^{81}\equiv1\pmod7

b^{91}\equiv2^{91}\pmod7

2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7
8\equiv1\pmod7
2\equiv2\pmod7
\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7

c^{27}\equiv6^{27}\pmod7

6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7

\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7
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Answer:

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Step-by-step explanation:

x+y=5  ==   -2x - 2y = -10

2x-y = 1 ==  2x - y = 1    +

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