4's only

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 🙂

and so

-sqrt(4)(log(log(sqrt(4))/log(4)))/log(4) = 1
-sqrt(4)(log(log(sqrt(sqrt(4)))/log(4)))/log(4) = 2
-sqrt(4)(log(log(sqrt(sqrt(sqrt(4))))/log(4)))/log(4) = 3
.
.
.
.

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.

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.

Originally posted by AThousandYoung
4^4+4/4

PEMDAS

No Parentheses
Exponenent is next...

256+4/4

Then Multiplication and Division...

256+1

Then Addition and Subtraction...

257

Parentheses are unnecessary to avoid ambiguity. Order of Operations...

Originally posted by sonhouse
Can you imagine what that solution would look like for 1,000,000,000,000🙂

Wait a minute, couldn't 1000 more easily be rendered thus:

4(4^4)-4!

Or has this been done already?

Originally posted by Soothfast
Wait a minute, couldn't 1000 more easily be rendered thus:

4(4^4)-4!

Or has this been done already?

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.

I think that I have got a long way towards 100. If anybody is interested I'll go and find my solution . . . . . . TWS

Originally posted by lemondrop
11 is open for a solution

Originally posted by coquette
For 11, I'll ask the "right-brained"question:

Why not take the already solved 13 and subtract the already solved 2? Trivial, no? Or am I missing something?

Originally posted by wolfgang59
You can only use [b]FOUR 4's[/b]

11 == 44/(sqrt(4) + sqrt(4))

=(4!-sqrt(4))*sqrt(4)/4
=-sqrt(4)(log(log(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(4))))))))))))/log(4)))/log(4)