logs are permissible I would assume
as long as it is a mathematical symbol, function, etc
very impressive the solution for 1000
I have no idea how to work that out
Originally posted by lemondrop logs are permissible I would assume
as long as it is a mathematical symbol, function, etc
very impressive the solution for 1000
I have no idea how to work that out
Using the fact that log(x^n) = n log x we have that (I think there is one too many square roots)
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 iamatiger Can anyone answer the following bonus question:
How many zeros do you need, combined with any other common mathematical symbols and constants, but no other digits, to make any integer?
Well 0 factorial gets you a 1, and if we require an integer greater than one to apply the technique you mentioned above, then noting that 0! + 0! gives you 2 then 3 * 2 = 6. Which is the answer I'm going to shoot for.
Originally posted by Agerg Well 0 factorial gets you a 1, and if we require an integer greater than one to apply the technique you mentioned above, then noting that 0! + 0! gives you 2 then 3 * 2 = 6. Which is the answer I'm going to shoot for.
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.