Q: How do you make 27 using 1 2 3 and 4 only once?

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In the case of 27, there is only one possible factor tree - using the factors 3 and 9 as the first step. 27 3 x 9 3 x (3x3)

27 can be written as 2.70 × 101

If you can repeat a digit, there are 27. If you can't repeat a digit, there are only 6.

The factors of 27 are 1, 3, 9, and 27.The factor pairs of 27 are 1 x 27 and 3 x 9.The proper factors of 27 are 1, 3, and 9 or,if the definition you are using excludes 1, they are 3 and 9.The prime factors of 27 are 3, 3, and 3.Note: There is repetition of these factors, so if the prime factors are being listed instead of the prime factorization, usually only the distinct prime factors are listed.The only distinct prime factor is 3. The prime factorization of 27 is 3 x 3 x 3 or, in index form (in other words, using exponents), 33.NOTE: There cannot be common factors, a greatest common factor, or a least common multiple because "common" refers to factors or multiples that two or more numbers have in common.

3+3+3x3=27

Related questions

[Impossible].

3^((4+1)/9)=35/9=243/9=27

(6*4)+2+1 = 27

3*(2*4 + 1) = 3*(8 + 1) = 3*9 = 27

Once only

Six if using each letter once, or 27 if using a letter more than once.

This is for a job application...but I think it's impossible. Any takers with an answer?

32 -(1+4) =27

8

9*9/3 + 6 - 2 = 9*3 + 6 - 2 = 27 + 6 - 2 = 31

It is: 5*(27-20)-5 = 30

4! + 1 x 6 Ã· 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 Ã· 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.