I have a solution for 107 if 0.4 recurring is allowed... (4! * sqrt (4) - 0.4(recurring)) / 0.4(recurring) = 107 If it's not allowed, I don't know what else could be done really. EDIT: (e^4-sqrt(4))/.4-4! = 107.495375083 It kinda works but needs rounding so not really...
I'll allow it, I have also been trying to get 107, but didn't succeed... cheety one again... It is getting harder... 4!*4+4squared - 4
Ah, I know a maybe cheaty way for 107! Depends on if adding !'s to a gamma is allowed... 44 / .4 - gamma (4)!! = 107 110 - (4 - 1)!!! = 107 110 - 3!!! = 107 110 - 3 * 1 = 107 110 - 3 = 107 110 - 3 = 107
Actually, that could probably just be one ! (so 44 / .4 - gamma (4)! = 107), because 3! = 3 * 2 * 1 = 6, so 3!! would be 3 * 1 = 3. 3!!! would be just 3 = 3, so that would completely remove that 3 * 1 step.
Maybe if you use math.floor(x) you can round it. Not really sure if that is allowed though, since it is not really a math thing but more used for programming
Floor is used in maths, I think, and is usually presented as . (you can't write that using Unicode though, I think...)
(4 + gamma (sqrt (4)))! – 4 * sqrt (4) = 112 Ugh I don't like using gamma all the time I need to stop but at this point it is quite hard to. It does get easier about here, as 120 is easily achievable using only two 4s- (4 + gamma (sqrt (4)))! Then you can just add and subtract the smaller numbers; gamma (sqrt (4)), sqrt (4), 4, gamma (4), 4!!, etc...
If there is a large enough gap, can we assume people are stuck and help / give hints, or should we wait?
44 / .4 + gamma (4)! = 113 110 + (4 - 1)!! = 113 110 + 3!! = 113 110 + 3 * 1 = 113 110 + 3 = 113 110 + 3 = 113
I'd say giving hints after a long gap is a good thing: especially becuase this seems to get exponentially harder 44/.4 + 4 I am getting the easy ones here...
