Exponents & Mods: What You Need to Know

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The discussion focuses on understanding modular arithmetic in the context of exponents and tetration, specifically regarding the expression ^n3, which represents a power tower of threes. Tetration is formally expressed as 3^{3^{...^3}} with n threes, and Knuth's up arrow notation is used to denote this operation. A participant suggests using induction to prove that ^n3 mod 10 equals 7 for n greater than 1, starting with the base case of 3^3. The proof involves showing that the pattern holds for subsequent powers. The conversation emphasizes the relationship between modular arithmetic and exponentiation in advanced mathematical contexts.
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Could somebody at least tell me about mods in exponents?
 
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Sam Anderson said:
Let 3^n denote 333... (n+1 threes in total)

Tetration already has a formalized expression.

^n3=3^{3^{...^3}} 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

^n3 = 3\uparrow\uparrow n

So what's your question?
 
Mentallic said:
Tetration already has a formalized expression.

^n3=3^{3^{...^3}} 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

^n3 = 3\uparrow\uparrow n

So what's your question?
Thanks for the quick reply, but I wasn't finished typing. I accidentally hit enter. Sorry.
 
Sam Anderson said:
Could somebody at least tell me about mods in exponents?

I could only guess what you're asking for, but if I were to guess, say you want to prove that ^n3 \mod 10 = 7, n>1 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 3^3 = 27 hence 27 \mod 10 = 7 and we can then show by calculation that 3^7 \mod 10 = 7 hence it must be true for all n.
 
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