$P=1^1+2^2+3^3+4^4+5^5+\ldots \ldots \ldots+48^{48}+49^{49}+50^{50}$. What is the remainder when $P$ is divided by 8 ?
A. 1
B. 3
C. 5
D. 6
E. 7
Is modular arithmetic, chinese remainder, or Fermat the solution?
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Please edit to include your efforts. – lulu Oct 16 '22 at 22:57
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Modular arithmetic. You can ignore all but one even number. – Thomas Andrews Oct 16 '22 at 22:57
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Please include your efforts. We are NOT a Homework Answering Service! – Mike Oct 16 '22 at 23:11
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@Mike this seems like a competition problem. – Kamal Saleh Oct 16 '22 at 23:15
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1In this case I'd say brute force is the best bet. It's pretty trivial to figure out what $even^{even}\pmod 8$ is and using Fermat it's pretty easy to figure out what $odd^{odd}\pmod 8$ is. – fleablood Oct 16 '22 at 23:18
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1Closely related: https://math.stackexchange.com/questions/923752/the-remainder-of-112233-dots9898-mod-4 also https://math.stackexchange.com/questions/325529/find-x-such-that-sum-k-12014-kk-equiv-x-pmod-10 and probably many more. – Gerry Myerson Oct 16 '22 at 23:35
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$(2k+1)^2=8n+1$ – Bob Dobbs Oct 17 '22 at 00:26
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Have you looked at those links, Trans? – Gerry Myerson Oct 18 '22 at 23:09
1 Answers
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Note that $n^n\equiv0 \pmod{8}$ for all even $n\geq 4$. Fermat gives $n^4\equiv 1\pmod{8}$ for odd $n$, so \begin{align*} n^n\equiv n\pmod{8} \quad\text{if $n\equiv 1\pmod{4}$} \\ n^n\equiv n^3\pmod{8} \quad\text{if $n\equiv 3\pmod{4}$}. \end{align*} Thus, \begin{align*} P &\equiv 2^2+\sum_{k=0}^{12} (4k+1)+\sum_{k=0}^{11} (4k+3)^3\\ &\equiv 4+(7\cdot 1+6\cdot 5)+(6\cdot 3+6\cdot (-1))\\ &\equiv 53 \equiv5\pmod{8}. \end{align*}
dromastyx
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