3 years ago

Help. Last question for now. Thx . Please write a proof.

Mathematics

1 answer:

pashok25 [27]3 years ago

8 0

Answer:

<1 = <2 because...

We know that <1 = 90. So <2 = 90 since <1 and <2 form a linear pair.

Step-by-step explanation:

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

Blizzard [7]

$\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}$

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

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

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

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

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

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

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3 years ago

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How do you do<br>8(4x+7)=56

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