Hard math problem!?

Question

I know the answer to this, it's not homework, but here's the problem: you have 4 fours. You can use them any way you want, as part of a fraction, squared, cubed, anything. You can use numbers like 44, which counts for 2 fours. You can use as many operations as you'd like, but you have no other numbers but 4. (Excluding exponents) Make it so that you find a way to use these in equations that equal the numbers 1-20. It's harder than it looks!

Answer 1

I assume you have to use *exactly* four 4s in each equation.

44 / 44 = 1
(4/4) + (4/4) = 2
sqrt(4 * 4) - 4/4 = 3
sqrt(4 + 4 + 4 + 4) = 4
sqrt(4 * 4) + 4/4 = 5
4 + sqrt(4) + 4 - 4 = 6
4 + sqrt(4) + 4/4 = 7
4 + sqrt(4) + sqrt(4) = 8
4 + 4 + 4/4 = 9
4 + 4 + sqrt(sqrt(4) + sqrt(4)) = 10
44 / sqrt(4 * 4) = 11
4 + 4 + sqrt(4) + sqrt(4) = 12
44 / 4 + sqrt(4) = 13
4 + 4 + 4 + sqrt(4) = 14
4 * 4 - 4/4 = 15
4 + 4 + 4 + 4 = 16
4 * 4 + 4/4 = 17
44 / sqrt(4) - 4 = 18
4! - 4 - 4/4 = 19
4² + sqrt(4) + 4/4 = 19
4 * 4 + sqrt(4) + sqrt(4) = 20

I'm not happy with either of my answers for 19. I don't like using the exponent ² because that feels like cheating using a 2. And the other choice is using the factorial symbol. I guess the intended answer is the one using 4².

Answer 2

Almost got them all, but I ran out of time.

44/44 = 1
(4/4) + (4/4) = 2
(4+4+4)/4 = 3
sq(4) - (4 + 4 + 4) = 4

(4 + 4) / 4 + 4 = 6
44/4 - 4 = 7
4 + 4 + 4 - 4 = 8
(4/4) + 4 + 4 = 9
(44 -4) / 4 = 10
4 + 4 + 4 - sqrt(4) = 11
4 + 4 + sqrt(4) + sqrt(4) = 12
(44 / 4) + sqrt(4) = 13
sqrt(4) + 4 + 4 +4 = 14
(44/4) + 4 = 15
((cube 4)/4) -4 + 4 = 16
(4 * 4) + (4 / 4) = 17
(4 * 4) + 4 - sqrt(4) = 18

((4 / 4) + 4) * 4 = 20

Answer 3

1. Let S be a set such that for each element x of S there exists a unique element x' of S. 2. There is an element in S, we shall call it 1, such that for every element x of S, 1 is not equal to x'. 3. If x and y are elements of S such that x' = y', then x = y. 4. If M is any subset of S such that 1 is an element of M, and for every element x of M, the element x' is also an element of M, then M = S. Just as a matter of notation, we write 1' = 2, 2' = 3, etc. We define addition in S as follows: (a1) x + 1 = x' (a2) x + y' = (x + y)' The element x + y is called the sum of x and y. Now to prove that 1 + 1 = 2. From (a1), with x = 1, we see that 1 + 1 = 1' = 2.

Answer 4

44/44 = 1
4/4 + 4/4 = 2
sqrt(4) +sqrt(4) -4/4 = 3
sqrt(4) +sqrt(4) +sqrt(4) -sqrt(4) = 4
sqrt(4)+sqrt(4) + 4/4 = 5
(4+4)/4 +4 = 6
44/4 - 4 =7
4*4/4 + 4 = 8
4 +4 + 4/4 = 9
(44-4)/4=10
44/4 +sqrt(4) = 13
44/4 +4 = 15
4+4+4+4 = 16
4*4 + 4/4 = 17
4*4 +4/sqrt(4) = 18
4*4 +sqrt(4)+sqrt(4) = 20

Answer 5

C'est vraiment difficile ta question.
Essayons de la resoudre: tu as 44*2=88
44/2=22
88+22=110
44/8=5(au tronc)
5*2=10 et tu l'ajoutes au 110 qui te donnes les 120 que tu voulais.

Answer 6

1: 4/4*4/4
2: 4/4+4/4
3: 4-4/4
4: 4/sqrt(4)+4/sqrt(4)
5: 4+sqrt(4)-4/4
6: 4+sqrt(4)+4-4
7: 44/4-4
8: 4*4/4+4
9: 4+4+4/4
10: (44-4)/4
11: 44/4+4-4
12: 44/4+4/4
13: 4^2-4+4/4
14: 4^2-sqrt(4)+4-4
15: 4*4-4/4
16: 4*4-4+4
17: 4*4+4/4
18: 4^2+sqrt(4)+4-4
19: 4^2+4-4/4
20: (4+4/4)*4