Funnily enough, it's still shorter than pi (π)
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Oh man. As a pi enthusiast myself, I just spent about 5 minutes googling how an integer could be longer than an irrational number, since this was a fact I hadn't heard before. Need coffeepic.twitter.com/QVradpcMSZ
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Proof by Induction that 6ᵐ always ends in 6 1) 6¹=6 2) Assuming 6ᵐ=10k+6 3) For m+1, 6ᵐ⁺¹=6(10k+6)=60k+36=10(6k+3)+6
Thanks. Twitter will use this to make your timeline better. UndoUndo
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Oh yes! A Googolplex is 10^(10^100). 6^(6^6) is already way larger than a googol (~10^36000 vs. 10^100) and there's three exponentiations still to come.
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Which means there's not enough space in the universe to do it. Even if you could write on every particle, that'll only give you about 10^90 digits!
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2^123456789987654321^123456789988654321^987654321123456789 +1 and same no. - 1 both are primes
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Also 2^2^n+1 is a prime or 2^2^n - 1 is a prime but both can be primes for certain nos between 1 to 10.
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Fermat's little theorem once again with giving us the last digit and nothing else.
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It's rather modular arithmetic than FLT. But it gives us as many trailing digits as we want...
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It has to be 6^(6^(6^(... your brackets are wrong
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