This is a conversion someone on SE made:
$$77777\equiv1\pmod{4}\implies77777^{77777}\equiv77777^1\equiv7\pmod{10}$$
But I don't understand how this is done?
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possible duplicate of Find the last digit of $77777^{77777}$ – Macavity Feb 19 '15 at 10:15
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@Macavity Not really duplicate. More of a follow-up question which could have been asked in a comment to the respective answer. – AlexR Feb 19 '15 at 11:32
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@AlexR Agreed, though the OP has a pattern of doing this. – Macavity Feb 19 '15 at 11:33
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@Macavity, is it wrong to have follow up questions? – Lebes Feb 19 '15 at 11:35
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Not really, though the appropriate place for a clarification on the same question / answer is not somewhere else. – Macavity Feb 19 '15 at 11:36
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1If I have small questions, on which I dont need that much clarifiation is the comments. Otherwise I always beleived a new question. – Lebes Feb 19 '15 at 11:37
2 Answers
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What was actually used is Euler's theorem: $$a^{\varphi(n)} \equiv 1 \pmod n$$ Where $\varphi$ is the totient function. $\varphi(10) = 4$ so $$77777^{a + 4k} \equiv 77777^a \pmod{10}\\ \Rightarrow 77777^{77777} \equiv 77777^{77777 \bmod 4} = 77777^1 \pmod{10}$$
AlexR
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1@Lebes Because we want to compute $77777^{77777} \pmod{10}$ so we need the value of $\varphi(10)$ so we can apply Euler's theorem. – AlexR Feb 19 '15 at 10:10
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this answer more directly answers the question, though the other approach is important to understand as well. – MichaelChirico Feb 19 '15 at 12:48
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Hint: $7^2\equiv -1\pmod{10}$.
No base change actually happens.
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Yes, I used $\pmod 2$ to give a little bit less away - and then I added the second sentence to hint at how the first might be used. – Henrik supports the community Feb 19 '15 at 10:05
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1,Can you show me a full explanation? It isnt homework, I am just trying to under modular arithemtic? +1 – Lebes Feb 19 '15 at 10:06
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I like Henrik's style (+1). @Lebes, why don't you continue this and calculate the remainders mod $10$ for the next few powers of $7$?! – Jyrki Lahtonen Feb 19 '15 at 10:07
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1@JyrkiLahtonen, I did and got: $$77777^{n} \equiv 77777^{n + 4} \pmod{10}$$ but I dont know what to do next/ – Lebes Feb 19 '15 at 10:13
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@JyrkiLahtonen: Thank you. That actually made me feel really good about my answer. – Henrik supports the community Feb 19 '15 at 10:14
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Rinse and repeat. $$7777^0\equiv 7777^4\equiv7777^8\ldots$$ All modulo 10. – Jyrki Lahtonen Feb 19 '15 at 10:15
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@Lebes: When you've realised that $77777^n\equiv 77777^{n+4}$ it should be easy to find a $k$ such that $77777^n\equiv 77777^m$ where $n\equiv m\pmod{k}$. – Henrik supports the community Feb 19 '15 at 10:17
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1@JyrkiLahtonen, so: $$77777^{n} \equiv 77777^{n + 4k}$$ for integer $k$. Let $n = 1$. $$77777 \equiv 77777^{1 + 4k}$$
Let $$77777 = 1 + 4k \rightarrow k = 19444$$
$$77777 \equiv 77777^{77777} \equiv 7$$ Then?
– Lebes Feb 19 '15 at 10:18 -
@Lebes: That's correct. To summarize: You noticed that the last digits of the sequence $7777^n$ repeats in a cycle of length four. Therefore the members with $n=7777$ and $n=1$ share the last digit. With $n=1$ the task of calculating it is trivial :-) – Jyrki Lahtonen Feb 19 '15 at 11:49