3*(2*4 + 1) = 3*(8 + 1) = 3*9 = 27
3^((4+1)/9)=35/9=243/9=27
[Impossible].
(6*4)+2+1 = 27
32 -(1+4) =27
To make 27 using the numbers 1, 2, 3, and 4 only once, you can use the following mathematical expression: (3 * 4) + (2 * 1) = 12 + 2 + 1 = 27. This equation follows the order of operations (multiplication before addition) to correctly arrive at the desired result of 27.
Using 1, 2, and 3, you can make 27 whole numbers.
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 ).
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