How do you get 20 by using 4 4's.by performing fundamental operations.?

Question

How do you get 20 by using 4 4's.by performing fundamental operations.?

Answer 1

chdfgj

Answer 2

4((4/4)+4))=20.

AM I RIGHT????????

Answer 3

{4+(4/4)}.4

Answer 4

(4/4 + 4) (4) = 20

4 x 4 + √4 + √4 ( do you allow this one ? ! )

Answer 5

(4+(4/4))*4........

Answer 6

Suggest 4*4 + 4 = 20

Answer 7

4*4+4

Answer 8

(4+4/4) x 4 = 20

Answer 9

((4/4) + 4)* 4 = 20

sqrt(4)+sqrt(4) + (4*4)

(4! -4) * (4/4)

sqrt(4*4) +(4*4)

((4 mod 4) + 4 ) * 4 = 20  

Answer 10

nice question

just do (4 + 4/4) * 4 = 20

Answer 11

  ( 4+4/4) into4= 20.

Answer 12

{4! is known as factorial 4 = 1*2*3*4 = 24}

1 = (4/4) * (4/4)

2 = (4/4) + (4/4)

3 = (4 + 4 +4 ) / 4

4 = (4/4) *  Ã¢ÂˆÂš[ 4*4] or 4! - [(4+4)/.4]

5 = (4/4) +  Ã¢ÂˆÂš[ 4*4]

6 = 4 + [(4 + 4) / 4]

7 = 4 + (4/4) +  Ã¢ÂˆÂš(4)

8 = [(4+4)/4] * 4

9 = [4* √(4)] + (4/4)

10 = [4* √(4)] + [4/  Ã¢ÂˆÂš(4)]

11 = [4! / √(4)] - (4/4)

12 = (4*4) - [ √(4*4)] or (4 + 4) + [ √(4*4)]

13 = (4*4) - [ √(4*4)]

14 = (4 * 4) - [4/ √(4)]

15 = (4 * 4) - (4/4)

16 = (4*4) * (4/4) or [(4*4)/.4] - 4!

17 = (4 * 4) + (4 /4)

18 = (4*4) + [4/ √(4)] or 4! + 4 - (4/.4)

19 = 4! - [4+{4/4)]

20 = 4! - [4* (4/4)]

21 = 4! + (4/4) - 4

22 = 4! + √(4) - √(4*4)

23 = 4! - [4/( √(4*4)]

24 = 4! * [4/( √(4*4)]

25 = 4! + [4/( √(4*4)]

26 = 4! + [ √(4)*(4/4)]

27 = 4! + [4 - (4/4)]

28 = 4! + [4 *(4/4)] OR 44 - (4*4)

29 = 4! + 4 + (4/4)

30 = 4! +4 + [4/ √(4)] OR 4! +(4/.4) - 4

31 = 4! +  Ã¢ÂˆÂš(4) + [ √(4)/.4]

32 = 4! + 4 +  Ã¢ÂˆÂš(4*4) or 4! +(4/.4) -  Ã¢ÂˆÂš(4)

33 = 4! +  4 + [ √(4)/.4]

34 = 4! +  4 + 4 + √(4)

35 = 4!+ (44/4)

36 = 4! + (4*4) - 4 OR 44 - [4 *√(4)]

37 = 4! + 4c + 4 + 4 (4c is complement of 4, which is 9-4=5)

38 = 4! + [4*4-√(4)]

39 = 44 - √(4)/.4

40 = 4! + [4 *√(4*4)]

You can go on experimenting like this with 4 fours!

AJM