I have this problem: $\frac{431^5 + 611}{27}$
I'm supposed to find the principal remainder by hand. But I have no idea how to start when $431$ has en exponent of $5$. Can someone please explain? Thanks.
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Ridertvis
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5Can you determine $431 \bmod {27}$? Can you determine $611 \bmod {27}$? – Robert Shore Sep 21 '22 at 21:17
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1@RobertShore Is 431 (mod 27) = 26 ? And 611 (mod 27) = 17 ? If it isn't then I don't know. – Ridertvis Sep 21 '22 at 21:22
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3@Ridertvis: Your arithmetic is correct. Now, what is $26^5 \operatorname{mod} 27$? – Dan Sep 21 '22 at 21:23
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4Think of it as $431 \equiv -1 \pmod {27}$. That makes the exponentiation much easier. – Robert Shore Sep 21 '22 at 21:27
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@RobertShore Can you explain that further. Please. – Ridertvis Sep 21 '22 at 21:29
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2@Dan 26 according to my calculations. – Ridertvis Sep 21 '22 at 21:31
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1@Ridertvis: Good. (And I hope you took a shortcut instead of working out $26^5 = 11881376$ by hand.) Now, what's $(26 + 17) \operatorname{mod} 27$? – Dan Sep 21 '22 at 21:33
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1@Dan If I have understood this correctly then that should be 16. – Ridertvis Sep 21 '22 at 21:35
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@Ridertvis: Yes, it's 16. And my work here is done. – Dan Sep 21 '22 at 21:43
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1The point is that $431 \equiv -1 \pmod{27} \Rightarrow 431^5 \equiv (-1)^5 = -1 \pmod {27}$. – Robert Shore Sep 22 '22 at 01:20
1 Answers
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We write the fraction as $\dfrac{(15\times27+27-1)^{5}+21\times27+44}{27}$
which is equal to $\dfrac{M\times\,27\,-1+44}{27}=M+\dfrac{43}{27}$
and $43=27+16$, hence the remainder is $16$.
I am sorry I have not checked the result with a calculator.
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I don't understand what happens between step 1 and 2. Where does big M come from? – Ridertvis Sep 21 '22 at 22:10
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1When we have $(A27-1)^{5}$ by the binomial identity this is a multiple of $27 -1$. So in the nominator we have a multiple of $27$$ -1+44=M27+43$ – Sep 21 '22 at 22:14