4's only

33 = 4! + 4 + sqrt(4)/.4
34 = 4! + 4 + 4 + sqrt(4)
35 = 4! +44/4
36 = 4! + (4*4 - 4)

37 = 4! + (4!+sqrt(4))/sqrt(4)
38 = 4! + 4*4 - sqrt(4)
39 = (4*4-.4)/.4
40 = 44 - sqrt(4) - sqrt(4)
41 = (4*4+.4)/.4
42 = 44 - 4 + sqrt(4)
43 = 44 - 4/4
44 = 44 - 4 + 4
45 = 44 + 4/4
46 = 44 + 4 - sqrt(4)

41 was nice
is there another solution for 41?

Not that I found, did you have a different solution for 39? ( My sols for 41 and 39 are similar)

no
good job

47 = 4!+4! - 4/4
48 = 4! + 4! + 4 - 4
49 = 4! + 4! + 4/4
50 = 4! + 4! + 4 - sqrt(4)

half way there - I'll let someone else have a go now

by the way, are we allowed log?

4/(log(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(4)))))))))/log(4))-4! = 1000

4!/.4 -4/.4... = 51 (where the ... after the last 4 denotes 4 with a bar on it - i.e. 4 recurring)
44 + 4 + 4 = 52
44 + 4/.4... = 53
44 + 4/.4 = 54
(4! - .4)/.4 - 4 = 55
4! + 4! + 4 + 4 = 56
(4! + .4)/.4 - 4 = 57
(4^4 - 4!)/4 = 58
4!/.4 - 4/4 = 59
44 + 4 * 4 = 60

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

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
.
.
.
.

The puzzle has been solved, I think

Yes, we can even make every integer with just 3 fours (or, with a few small modifications, 3 of almost anything, except -1, 0 and 1):

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

and so on
The bit in bold changes, it has n nested square roots, where n is the integer required.

It works because:
sqrt(x) = x^(2^-1)
sqrt(sqrt(x)) = x^(2^-2)
sqrt(sqrt(sqrt(x))) = x^(2^-3)

the maths surrounding the nested sqrts is needed to extract the bold number inside the power of x.

the technique doesn't work for 1 or 0 because sqrt(1) = 1 and sqrt(0) = 0, so the nesting of the sqrts has no effect at all.

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?

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?

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.