+0  
 
0
119
41
avatar+73 

I have an assignment requiring me to make 1-100 using 2,0,2,2 and I have done the first 32.

You can use +-x/ square root decimal point power factorial double factorial () permutation and combination.

The deadline of the assignment is 7th May 15:59:59 UTC

I am stuck,can anybody help me please? Thank you.

 Apr 23, 2021
 #1
avatar+121003 
0

Can  we   combine digits   ???....such as 2, 0  =   20  ????

 

 

cool cool cool

 Apr 23, 2021
 #3
avatar+73 
0

Yes

but no concatenation like (2-1)0=10

Matherew  Apr 23, 2021
 #2
avatar+73 
+1

1=2x0+2/2

2=-20+22

3=2x0!/2+2

4=2x0+2+2

5=2-0+2+2

6=2+0+2+2

7=2+0!+2+2

8=2x2x2+0

9=2x2x2+0!

10=sqrt(20/2)^2

11=(2+2)!/2-0!

12=(2+2)!/2-0

13=(2+2)!/2+0!

14=(2+2)!!+2+0!

15=2^(2x2)-0!

16=2^(2+2)+0

17=2^(2+2)+0!

18=(2+2)!-(2+0!)!

19=22-2-0!

20=-2-0+22

21=-2+0!+22

22=2x0+22

23=(2+2)!-2+0!

24=2-0+22

25=(2+2)!+2-0!

26=(2+2)!+2+0

28=22+(2+0!)!

29=(.2)^(-0!)+(2+2)!

30=(2+2)!+(2+0!)!

31=(.2)^(-2)+(2+0!)!

32=2^(2+2+0!)

 Apr 23, 2021
 #4
avatar+121003 
+2

Here's a few  ....maybe I can think of some others

 

[.2^-2 ]  *2 - 0!   =   49

[.2^-2] *  2  + 0!   = 51

[.2^-2] * 2  / 0!  =  50

 

C (20, 2)  / 2  =    95

 

(2 + 2)!  *  ( 2!  + 0!)   =  72

 

(2  + 2)!  +  20  =   44

 

(.2^-2)  + 20  = 45

 

[ 2^(2!  + 0!)  ] ^2  =  64

 

20  + 22   = 42

 

 

cool cool cool

 Apr 23, 2021
edited by CPhill  Apr 23, 2021
 #5
avatar+73 
+1

Literally I already know about 4 of them

But still, thx

Matherew  Apr 23, 2021
 #30
avatar+73 
0

16 to go

Matherew  Apr 25, 2021
 #6
avatar+121003 
+1

(20 - 2 ] * 2  =  36

 

[ 22 * 2 ]  - 0!  =  43 

 

[ 2 (20) ]  - 2   =  38

 

[ .2^--2  + 0! ]  *  2  =   52 

 

[.2^-2  - 0!  ] * 2   =  48

 

(20/2)^2   =  100

 

 

 

cool cool cool

 Apr 23, 2021
edited by CPhill  Apr 23, 2021
 #7
avatar+73 
+1

Do you have a program or sth like that?

Matherew  Apr 23, 2021
 #8
avatar+73 
0

I've done 52 and 48 but you seems like overkilled 48

Matherew  Apr 23, 2021
 #9
avatar+121003 
+1

No program....just off the  top of my head.....

 

I did something  like this  before.....I don't believe  that  either  me, the questioner (or his teacher)  could find all of them....I think we  found something like  83  of them

 

I'll see if  I can find the post....

 

 

cool cool cool

CPhill  Apr 23, 2021
 #10
avatar+73 
+1

EPIC

Maybe we should find someone who is good at programming (pun intended)

Matherew  Apr 23, 2021
 #11
avatar+121003 
0

LOL!!!

 

I'll  see if I can come up with some more tomorrow.....look under  this post and  check

 

cool cool cool

CPhill  Apr 23, 2021
 #12
avatar+73 
0

Okay 

Matherew  Apr 23, 2021
 #20
avatar+73 
0

Btw, which time zone r u in?

Matherew  Apr 23, 2021
 #22
avatar+73 
0

Did u think of any other?

Matherew  Apr 24, 2021
 #29
avatar+73 
0

Any other ideas?

Matherew  Apr 25, 2021
 #31
avatar+73 
0

Can u think of some of the 80s?

Matherew  Apr 25, 2021
 #13
avatar+73 
0

33=(.2)th root(2)+2-0!

34=(.2)th root(2)+2-0

35=(.2)th root(2)+2+0!

36=(20-2)x2

 

33-36 cleared

 Apr 23, 2021
edited by Matherew  Apr 23, 2021
 #14
avatar+73 
0

37=.2th root(2)+(.2)^(-0!)

38=.2th root(2)+h2c0!)!

 

 

37 and 38 cleared

 Apr 23, 2021
 #15
avatar+73 
0

Update

39,41,54-100 left

 Apr 23, 2021
edited by Matherew  Apr 23, 2021
edited by Matherew  Apr 23, 2021
 #16
avatar+73 
0

41,67-100 left

 Apr 23, 2021
edited by Matherew  Apr 23, 2021
 #17
avatar+73 
0

80% done

 Apr 23, 2021
 #18
avatar+73 
0

Remaining

41,54,67-69,71,74,79,82-89,91,93,97

 Apr 23, 2021
 #19
avatar+73 
0

Sorry, I missed 70

Matherew  Apr 23, 2021
 #21
avatar+73 
0

Remaining:21

41 54 69-71 74 79 82-89 91 93 97

 Apr 24, 2021
 #23
avatar+73 
0

Tools

5=Sqrt((.2)^(-2))+0!

48=((2+0!)!)!!

8=(2+2)!!

100=(.(0!))^(-2)

 Apr 24, 2021
edited by Matherew  Apr 24, 2021
 #24
avatar+73 
0

Tools

5=Sqrt((.2)^(-2))=(.2)^(-0!)

48=((2+0!)!)!!

8=(2+2)!!

100=(.(0!))^(-2)

6=(2+0!)!

 Apr 24, 2021
 #25
avatar+73 
0

84%done

 Apr 25, 2021
 #26
avatar+73 
0

Maybe we can use infinite square roots

 Apr 25, 2021
 #27
avatar+73 
0

Remaining

67 to 69

71

74

77

79

82 to 84

86 to 89

91

93

 Apr 25, 2021
 #28
avatar+73 
0

Tools

5=Sqrt((.2)^(-2))=(.2)^(-0!)

48=((2+0!)!)!!

8=(2+2)!!

100=(.(0!))^(-2)

6=(2+0!)!

15=5!!

 Apr 25, 2021
 #32
avatar+73 
0

15 Remaining

67

69

71

74

77

79

82 to 84

86 to 89

91

93

 Apr 25, 2021
 #33
avatar+73 
0

We can use infinite nested square roots

 Apr 25, 2021
 #34
avatar+121003 
+1

Allowing for  the left factorial

 

! [ 2 * 2 ] ^2  +  0!   =   ! [ 4]^2 + 1  =   [9]^2  + 1  =   82

 

! [ 2  + 2  +  0! ] *  2   = ! [ 5 ] *2  =   [44] * 2  =   88

 

([ 2 + 0! ] !) !!   +  20  =  [ [ 3! ]!!  + 20  =  6!! + 20  = 48 + 20  = 68

 

C ( !(2 + 2) , 2 + 0!)  =  C ( 9,3)  = 84

 

 

cool cool cool

 Apr 25, 2021
 #35
avatar+73 
0

I need to ask if that's ok or not

Matherew  Apr 25, 2021
 #37
avatar+73 
0

82=((2+2)!!)/(.(0!))+2

Matherew  Apr 25, 2021
 #38
avatar+73 
0

That's not allowed ,just asked

Matherew  Apr 25, 2021
 #36
avatar+73 
0

10 remaining

67

69

83

84

86 to 89

91

93

 Apr 25, 2021
 #39
avatar+73 
+1

7 remaining

67

69

83

84

86 

87

93

 Apr 25, 2021
 #40
avatar+121003 
0

Wow...that's pretty  amazing  !!!!

 

cool cool cool

CPhill  Apr 25, 2021
 #41
avatar+73 
+1

I am starting to doubt that there are no solutions for the seven remaining

 Apr 26, 2021

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