3^((4+1)/9)=35/9=243/9=27
[Impossible].
(6*4)+2+1 = 27
32 -(1+4) =27
Oh, dude, this is like a brain teaser, but I got you. So, if you take 3 and multiply it by 4, you get 12. Then, add 1 to that, giving you 13. Finally, just cube 2 (which is 8) and add that to 13, and bam, you've got 27. Math can be fun, right?
To make 27 using the numbers 1, 2, 3, and 4, you can use the following expression: ( (4 \times 3 \times 2 \times 3) + 1 = 27 ). Here, you multiply 4, 3, and 2 to get 24, then multiply by another 3 to reach 72, and finally add 1 to get 27. Alternatively, you can also achieve 27 by using exponentiation, for example, ( 3^3 = 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