The editors should shut up if they don't know how.
Yes, if only +-*/() is allowed, It would be impossible.
If recurring digits is allowed, The answer is:
4! / (.4` * sqrt(.4`)) - 4
While .4' being 0.4444444444...
23
4-4*4+4=4
4+4-(4/4)=7
There are 100,230 fours.There are 100,230 fours.There are 100,230 fours.There are 100,230 fours.
It means that a 4 piston engine is smoothly running because it's 'firing on all fours'
23
To achieve the number 77 with four fours, the solution is as follows: sqrt(4/.4)4 - 4 Instead of the second four being equal to four tenths, place a dot over the top to indicate it being equal to four ninths.
There are 77 eighty-fours in 672
4+4(4-4)
101
4 /5
Using four fours, you can get 19 with the following sum: 4! - 4 - (4 / 4) That is, four factorial minus four - (four divided by four) - 24 - 4 - (4 / 4) = 19
(4!/4)4 - 4
44 + 4/4
To get negative 2 using four fours, you can use the following expression: (4 - 4) / 4 - 4 = -2. This equation breaks down as follows: subtracting 4 from 4 gives 0, then dividing by 4 gives 0, and finally subtracting 4 gives -4.
(4+4)/4 + 4 =6
-4-4+4/4