Challenge of the four!

Challenge of the four!

Using always 4 times the number 4 making use of any mathematical operation known, we must find the values.

As an example I made up the number 11. The answers are not unique!

Does someone to finish the challenge?

1 = (4 / 4) x (4 / 4)
2 = (4 x 4) / (4 x 4)
3 = (4 x 4 - 4) / 4
4 = (4 - 4) x 4 + 4
5 = (4 x 4 + 4) / 4
6 = (4 + 4) / 4 + 4
7 = 4 + 4 – (4 / 4)
8 = 4 x 4 – (4 + 4)
9 = 4 + 4 + (4 / 4)
10 = (44 – 4) / 4
11 = [44 / Sqrt(4)] / Sqrt(4)

12 =
13 =
14 =
15 =
16 =
17 =
18 =
19 =
20 =
21 =
22 =
23 =
24 =
25 =
26 =
27 =
28 = (4+4)*4-4
29 = 4!+4+(4/4)
30 =
31 =
32 =
33 =
34 =
35 =
36 =
37 =
38 =
39 =
40 =
41 =
42 =
43 =
44 =
45 =
46 =
47 =
48 =
49 =
50 =

PS: I'll answer a random number to each day that passes with no answer to the challenge.

Quote from Oberster Hirte

Well … thanks for this riddle;) But I don't have a clue about 33, 39 and 41 …

This numbers are the hardest one!

37 = 4!+(4!+Sqrt(4))/Sqrt(4)
37 = 24 + (24 + 2) / 2
37 = 24 + 26 /2
37 = 24 + 12
37 = 36

I think it's wrong …

By the way, did you know anything about (4?) operation?

It's like the 4! = (4 * 3 * 2 * 1)
4? = (4 + 3 + 2 + 1) - I've saw it at a book …

Maybe now I can do it … And the others numbers too.

Quote from QMP Night

Ähm... Evandro... 26/2 is in fact not 12 but 13...
37 = 4!+(4!+Sqrt(4))/Sqrt(4)
37 = 24 + (24 + 2) / 2
37 = 24 + 26 /2
37 = 24 + 13
37 = 37

Yes...

Sorry!