3*(2*4 + 1) = 3*(8 + 1) = 3*9 = 27
34/(1 + 2)
3^((4+1)/9)=35/9=243/9=27
[Impossible].
(6*4)+2+1 = 27
32 -(1+4) =27
Using 1, 2, and 3, you can make 27 whole numbers.
13 + 27 + 54 + 6 = 100.
This is for a job application...but I think it's impossible. Any takers with an answer?
4! + 1 x 6 ÷ 2 is one way I can figure without using an operation more than once. 4! = 4 x 3 x 2 x 1 = 24. Then by order of operations, perform 1 x 6 ÷ 2 = 3, which is added to 24 to get 27.Another way is: (6/2)^(4-1), which is 3 raised to the 3 power (3 x 3 x 3 = 27).If the limitation on operators is not present, then you can do it by 4 x 6 + 1 + 2 : 4 x 6 = 24, then add 1 and 2 to get 27.
8
4*32 - 1 = 4*9 - 1 = 36 - 1 = 35
write the largest number you can make using each of the digits 7,1,0,2, and 9 just once