Originally posted by Agerg Well if we're allowed logs then (as per the wikipedia article for this puzzle) we have
-sqrt(4)log(log(sqrt( ... n times ... (sqrt(4) ... ))/log(4))/log(4)
= -2 log(2^(-n)log(4)/log(4))/log(4)
= 2n log(2)/log(4) = 2n * .5 = n 🙂
Originally posted by sonhouse How does 0! = 1? I thought factorial for zero would be zero, zero times one should be zero and one times zero should be zero.
How does 0! = 1? I thought factorial for zero would be zero, zero times one should be zero and one times zero should be zero. It is defined to be 1. A nice explanation of why can be found at:
Can you imagine what that solution would look like for 1,000,000,000,000🙂 Heh...I'll be honest and say that I cannot imagine what the solution would look like!
Hmm, following the pattern and using exactly 4 4's
"-(log(log(" & { "sqrt(" x 10^12} & "4)" & { " )" x 10^12} & "/log(4))/log(4/sqrt(4))"
Where & means concatenate the strings, and { string x n} means repeat string n times, I think that is 7000000000032 characters
according to http://amazingbibletimeline.com/bible_questions/q10_bible_facts_statistics/
there are 3,116,480 letters in a king james bible, so we could print the sum in the equivalent of 2,246,124 bibles. If each book was 4 cm thick, the stack of books would be 90km high.
Factorial n is the number of ways of arranging n different things, so the way I always think about it is that there is 1 way of having absolutely nothing.
But the log function works mindlessly for any integer, that was the point of it, not to give the most elegant answer for all integers, but to prove an answer exists for *every* integer.
Originally posted by iamatiger But the log function works mindlessly for any integer, that was the point of it, not to give the most elegant answer for all integers, but to prove an answer exists for *every* integer.
4's only
29 Jun '14 15:56
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