Thefour digets problem?

Question

Using the four digits problem, i need answers 1 through 50.?

Of 1, 2, 3, and 4 each digit is only used once. you may use addition, subtraction, multiplication (not division), exponents, and parentheses in any way you want to. Also you can use two digits to make one number, such as 12 or 34...i need this help!

Answer 1

I'll do the first half of them for you.

1^(2*3*4) = 1

1*2*3-4 = 2

3*1^(2*4) = 3

1*2³-4 = 4

4+1^(2*3) = 5

2*3*1^4 = 6

2*3 + 1^4 = 7

4*2*1³ = 8

1*2+3+4 = 9

1+2+3+4 = 10

1+2*3+4 = 11

4*3*1² = 12

4*3+1² = 13

4*3+2*1 = 14

4*3+2+1 = 15

(1^3)(2^4) = 16

2^4 + 1^3 = 17

2^4+3-1 = 18

2^4 + 3^1 = 19

4²+3+1 = 20

13 + 2*4 = 21

34 - 12 = 22

2*3*4-1 = 23

1*2*3*4 = 24

1 + 2*3*4 = 25

14 + 3*4 = 26

Hope this helps you!

Answer 2

2-1=1

4-2=2

4-1=3

4*1=4

3+2=5

3+2+1=6

4+3=7

2*4=8

2+3+4=9

(2*3)+4=10

(4*3)-1=11

4*3=12

(4*3)+1=13

(3+4)*2=14

3 * (4+1)=15

4^2 = 16

4^2 + 1 = 17

21-3=18

14 + 3 + 2 = 19

(4 ^2)+3+1=20

(2+3)*4+1=21

23-1=22

24-1=23

2*3*4=24

(2*3*4)+1=25

23+4-1=26

23+4=27

32-4=28

31-2=29

(2*3)*(4+1)=30

32-1=31

34-2=32

34-1=33

34 = 34

((3^2)*4)-1=35

3^2 *4 = 36

13+24=37

42-1-3=38

42-3=39

42-3+1=40

42-1=41

43-1=42

42+1=43

41+3=44

42+3=45

12+34=46

(4^2)+31=47

(4+3)^2-1=48

(4+3)^2=49

(4+3)^2+1=50