Q: How do you get 13 using only five two's?

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(22 + 2*2)/2

4! - 44/4 = 24 - 11 = 13

[(4*4)+(4*4)+(4*4)+4] / 4 = 13

4444=5...? using the basic arithmetic operations like +,*,-,/. but use only once each these symbols try it

13, 26, 39, 52, 65.

Related questions

23 + 2 19 + 3 + 3 17 + 5 + 3 17 and 4 twos 17, 2 threes and 2 twos 13 + 7 + 5 13 and 6 twos 13 and 4 threes 11 and 2 sevens 11 and 7 twos and a bunch more up to 10 twos and 5 11 twos and 3

The factors of 34 are 1, 2, 17, 34 To get to 13 you could use 13 ones or 6 twos and 1 one or some other combination of ones and twos.

(22 + 2*2)/2

There are a few options for listing five prime numbers using the digits zero to nine only once:2, 3, 5, 7, 8649012, 5, 13, 647, 809

No, there are only five atoms in freon-13

(4x4)+(4/4)-4 The above uses five 4's. An answer, using factorials, and only four is: (4!+4-sqrt(4))/sqrt(4)

Five

13

Only if you subtract three.

8 plus 1 , minus 5, plus 7 + 13

im only 13

4! - 44/4 = 24 - 11 = 13