tough trouble of modular math

Can you show me just how to confirm this : :

$$66^{11 k+66} + 11^{101 k+55}\equiv \left\{\begin{array}{ll} 22 \pmod{165}& \text{if $k$ is odd,}\\ 77 \pmod{165} & \text{if $k$ is even.} \end{array}\right.$$

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2019-12-02 02:54:13
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Answers: 3

First, $66^2 \equiv 66 \pmod{165}$. Consequently $66^{11k+66} \equiv 66 \pmod{165}$. Second, $11^3 \equiv 11 \pmod{165}$. Consequently for also $n > 0$, $11^n \equiv 121 \pmod{165}$, whereas for weird $n > 0$, $11^n \equiv 11 \pmod{165}$. If $n=101k+55$ after that $n$ is weird iff $k$ is also. Hence for weird $k$ we get $66+121 \equiv 22 \pmod{165}$, and also for also $k$ we get $66+11 \equiv 77 \pmod{165}$.

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2019-12-03 04:24:08
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Since $165=3\times 5\times 11$, job modulo each of $3$, $5$, and also $11$ independently and afterwards incorporate them making use of the Chinese Remainder Theorem.

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2019-12-03 04:24:05
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It is $\rm \equiv 0\ (mod\ 11)\:,\: $ and also $\rm\: mod\ 15\ $ it is $\rm\ 6^n \equiv 6\ $ plus $\rm\ (-4)^n \equiv -4,\:1,-4\:, 1\:, \ldots\ $ amounts to $\rm\ 2, 7, 2, 7\:\ldots$

Hence it is $\rm\ 22,77,22,77\:\ldots\ (mod\ 11*15)$

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2019-12-03 04:24:00
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