[√(4/.4*)]^4 - 4 = [√9]^4 - 4 = 3^4 - 4 = 81 - 4 = 77
N.B. the denominator within the square root is meant to be 0.4 recurring, which is written with a dot above the 4 but I am not able to do this.
0.44444...... = 4/9
Also, if you do (4-40)3 - 4
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
(4x4) - (4/4)
A solution to the four fours problem for the number 33 is: 33 = (4-.4)/.4+4!
11 = 4/.4 + 4/4
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
23
(4x4) - (4/4)
4 × ((4 ÷ 4) + 4)=20
A solution to the four fours problem for the number 33 is: 33 = (4-.4)/.4+4!
(4+4)/4 + 4
you can not
One way to use four fours to make 25 is 4 + 4 + 4 + √4, which equals 25.
4+4-(4/4)=7
11 = 4/.4 + 4/4
4 + 4 + (4/4) = 9