0 = [4 + 4 - (4 + 4)]
1 = [(4 + 4) / (4 + 4)]
2 = [4 * 4 / (4 + 4)]
3 = [(4 + 4 + 4) / 4]
4 = [4 + 4 - √4 - √4]
5 = [√4 + √4 +(4/4]
6 = [4 + √4 + 4 - 4]
7 = [4 * √4 - 4/4]
8 = [4 + 4 + 4 - 4]
9 = [4 + 4 + {4/4}]
10 = [44 - 4/4]
11 = [4!/√4 - 4/4]
12 = [4 * 4 - √4 - √4]
13 = [4! / {√4 + 4/4]
14 = [4 + 4 + 4 + √4]
15 = [4 * 4 - 4/4]
16 = [4 * 4 + 4 - 4]
17 = [4 * 4 + 4/4]
18 = [4 * 4 + 4 - √4]
19 = [4! - 4 - 4/4]
20 = [{4 * 4} + √4 + √4]
Sorry if u wanted an answer not using ! or √ but I haven't figured them out yet though I only need 4 more which are 13, 14 ,18 and 19 if u know the equations for any one of them then please tell me. And also if you need others not including ! or √ which r between 0 to 20 than tell me I'll tell ya
There are 625 of them - too many to list.
If you want 4-digit numbers, there are 24 of them.
Yes. 6 * 4 = 24 (4 * 4) + 8 = 24
term
Technically, you cannot find a median if you have only two numbers, so it would be more reasonable to use an average instead.
22x2x2 is 88
how do i make 17 using only 2 4 6 8
2+10+6+4
Using only sums and differences, and not necessarily all four numbers, 1, 3, 9 and 27 will make all numbers from 0 to 40.
You can't unless something is missing in your question.
Four five nine
1,3,4,4
A bed - it has one foot (the foot of the bed) and four legs (which make it stand as a bed)
You can do just about anything if you use 4/4 as one of your addends.
17 is the only prime number that is the sum of four consecutive prime numbers. 2 + 3 + 5 + 7 = 17
Because there are only four numbers that divide into those numbers evenly.
the are only four