_ (4/0.4)^sqrt4-4 =77 9 81 =77
(4+4)*(4+4)=8*8=64
(44 - 4)/4
77 + 7/7 = 78
The only solution that I have ever seen uses some notation that is not so commonly used, namely a bar over 4 to mean the infinitely repeating decimal .44444... which is 4/9 . So, if x = 4/9 is written with one four, then sqrt(4/x) 4 - 4 uses four fours and equals 77 since 4/x = 4 * 9/4 = 9 and sqrt(9)4 - 4 = 34 - 4 = 81 - 4 = 77.
_ (4/0.4)^sqrt4-4 =77 9 81 =77
The answer is 4! - [(4+4)/4]
(4x4)-(4/4) = 15
4! - sqrt 4 + 4/4
4+4+4x4 by the power of 0.
(4+4)*(4+4)=8*8=64
4/.4 + 4/4 [= 10+1 = 11]
(44 - 4)/4
44/4 + 4! = 11 + (4x3x2x1) = 11 + 24 = 35
4! - sqrt(4)*sqrt(4)/4 = 23
how do i make 17 using only 2 4 6 8
77 + 7/7 = 78