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Could somebody at least tell me about mods in exponents?

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- Thread starter Sam Anderson
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Could somebody at least tell me about mods in exponents?

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Mentallic

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Tetration already has a formalized expression.Let 3^{^n}denote 3^{33... (n+1 threes in total)}

[tex]^n3=3^{3^{...^3}}[/tex] with n 3's in the stack.

Or you can even use Knuth's up arrow notation. Exponentiation is denoted by 1 up arrow, and tetration by 2, so

[tex]^n3 = 3\uparrow\uparrow n[/tex]

So what's your question?

- #3

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Thanks for the quick reply, but I wasn't finished typing. I accidentally hit enter. Sorry.Tetration already has a formalized expression.

[tex]^n3=3^{3^{...^3}}[/tex] with n 3's in the stack.

Or you can even use Knuth's up arrow notation. Exponentiation is denoted by 1 up arrow, and tetration by 2, so

[tex]^n3 = 3\uparrow\uparrow n[/tex]

So what's your question?

- #4

Mentallic

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I could only guess what you're asking for, but if I were to guess, say you want to prove that [itex]^n3 \mod 10 = 7, n>1[/itex] we would do an inductive proof. I'll make it short and you can fill in the details, but essentially you start with the base case [itex]3^3 = 27[/itex] hence [itex]27 \mod 10 = 7[/itex] and we can then show by calculation that [itex]3^7 \mod 10 = 7[/itex] hence it must be true for all n.Could somebody at least tell me about mods in exponents?

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