(4 x 4 x √4) - √4 = 32 - 2 = 30
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(4/4)*4!/4 = 6
(44 - 4)/4 x 4/4
4*4!/(4+4) = 4*24/8 = 12
4! - 44/4 = 24 - 11 = 13
444 + 444 + 44 + 44 + 4 + 4 +4*4
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
4*4*4-(4/4)
(4x4+4)/4
4/4+4/4=2
I am not aware of a solution using the four basic operations of arithmetic, but otherwise: 129 = [(4^4)/sqrt(4)] + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...))) * * * * * * * * * * * * No matter how many times you square root it, it will not equal 1 To the OP. I have spend a lot of time on this one. Just wanting to make sure that the correct number is 129 and you are sure you are using only 4 fours. It would work out great if using 5 fours. If you made a mistake, correct it and I'll check back. In the mean time, I'll continue thinking about a solution for 4 fours and 129.
(4+4)/4 + 4 =6
4 x 4 + 4/4 = 17
4-4*4+4=4
A solution to the four fours problem for the number 33 is: 33 = (4-.4)/.4+4!
4+4+4=12 12/4=3
4/4 + 4 × 4/4