2013 Math Game

[walker_boh_65]

Each year a few of my friends play a math game in which we try to use the the digits in the year (this year being 2013) to create the numbers 1-100 (This is not a game we created we found it years ago). I thought I would share it with you guys because I know at least few of you enjoy maths and logic problems.

The rules are pretty simple:
Use each digit once, and only once.
The following mathematical functions are accepted:

Spoiler

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addition
subtraction
multiplication
division
grouping symbols
decimals
raised to a power
square roots
factorial (x! ex: 5! = 5*4*3*2*1 )
double factorial (x!! ex: 9!! = 9*7*5*3*1 6!! = 6*4*2

EDIT: You can also combine to digits. ex. 20 or 31

Obviously the winner is the one who can make the most numbers from the 4 digits. We decide the winner amongst us at the end of the month.

If you would like to post any solutions, I ask that you please do so in a spoiler.

Have fun and good luck.

[Crazylemon64]

Ooh, puzzles!

I got the first 33, I don't want to spend too much time on this though.

My Solutions

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1
2
3
4 (1+3)
5 (3+2)
6 (2*3)
7 ((2*3) + 1)
8 ((3+1) * 2)
9 (3^2)
10 (1&0)
11 (10 + (3-2))
12 (10+2)
13 (10+3)
14 ((2&0)-3!)
15 (10+(2+3))
16 (2^(1+3))
17 ((2&0)-3)
18 ((2&0)-(3-1))
19 ((2&0)-1)
20 (2&0)
21 (2&1)
22 ((2&0)+(3-1))
23 (2&3)
24 ((2&0)+(3+1))
25 (((3!)-1)^2)
26 ((1&3)*2)
27 (3^(2+1))
28 ((3&0)-2)
29 ((3&0)-1)
30 (3&0)
31 (3&1)
32 (3&2)
33 (3&(1+2))

[dawnraider]

Being extremely interested in math, I may have to try this out with my friends. Also, it should be accepted not excepted, as right now you are stating that none of those are allowed :)

[Graphite]

Ooh, puzzles!

I got the first 33, I don't want to spend too much time on this though.

My experience with puzzles like these are that all digits MUST be used exactly once, in which case you would have to rethink some of those solutions. 3 would have to be (2^0 - 1) + 3, for example.

[Mason11987]

Not gonna lie, pretty proud of my 34. Also used every digit. Stuck on 43

Spoiler

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1 = (2+0+1)/3
2 = 3+1-2+0
3 = 3^0+2*1
4 = 3^0+2+1
5 = 3+0+2*1
6 = 3+0+2+1
7 = 3*2+1+0
8 = 3^2-1+0
9 = 3^2*1+0
10 = 3^2+1+0
11 = 10+3-2
12 = 12+3*0
13 = 13+2*0
14 = 13+2^0
15 = 13+2+0
16 = 10+3*2
17 = 20-3*1
18 = 20-3+1
19 = 20-1^3
20 = 21-3^0
21 = 21+3*0
22 = 21+3^0
23 = 23+1*0
24 = 23+1^0
25 = (3!-1)^2
26 = 13*2+0
27 = 3^(2+1)
28 = 30-2^1
29 = 30-1^2
30 = 30*1^2
31 = 30+1^2
32 = 30+2^1
33 = 30+2+1
34 = 32+1+0!
35 = 3!^2-1+0
36 = 3!^2+1*0
37 = 3!^2+1+0
38 = 3!^2+1+0!
39 = 13*(2+0!)
40 = 20*(3-1)
41 = (Messed this up)
42 = 21*(3-0!)
43 =

Crazy used: 33 (3&(1+2))

I didn't use the join after calculating thing, seemed wrong to me :).

[Graphite]

Just realised something. Walker's simply letting us solve the puzzle for him ;)

[PatrickSJ]

So far I have 63 of the 100 numbers although it is not in sequence.

Spoiler

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1 3-2+(1*0)
2 3-1+(2*0)
3 3*1+(2*0)
4 3+1+(2*0)
5 3+2+(1*0)
6 3*2+(1*0)
7 3*2+1+0
8 (3+1)*2+0
9 3*(1+2)+0
10 30/(1+2)
11 2*3!-1+0
12 2*3!+(1*0)
13 12+3^0
14 13+2^0
15 10+3+2
16 (3+2)!!+1+0
17 (3+2)!!+1+0!
18 2^3+10
19 20-(1^3)
20 20+(3*0)
21 21+(3*0)
22 21+(3^0)
23 3+(2*10)
24 21+3+0
25 21+3+0!
26 13*2+0
27 13*2+0!
28 30-(2^1)
29 30-(1^2)
30 3!*10/2
31 31+(2*0)
32 32+(1*0)
33 31+2+0
34 3!^2-1-0!
35 3!^2-1+0
36 3!^2+(1*0)
37 3!^2+1+0
38 3!^2+1+0!
39 13*(2+0!)
40 20*(3-1)
41
42 12+30
43
44 (3+1)!+20
45
46 (3*2)!!-1-0!
47 (3*2)!!-1+0
48 (3+2+1)!!+0
49 (3*2)!!+1+0
50 (3+2)*10
51 21+30
52
53
54
55
56
57
58 3!*10-2
59 2*30-1
60 2*30*1
61 2*30+1
62 31*2+0
63 21*3+0
64 21*3+0!
65 130/2
66
67 201/3
68
69
70 210/3
71
72 (3*2)!/10
73
74
75
76
77
78
79
80 (2^3)*10
81
82
83
84
85
86
87
88
89
90 (3^2)*10
91
92
93
94
95
96 sqrt((2^10))*3
97 (10^2)-3
98
99
100

Edit: Added solutions for 65, 67, and 70.

[dawnraider]

Just realised something. Walker's simply letting us solve the puzzle for him ;)

The best kind of slaves are those that think they are doing the work of their own free will :)

[walker_boh_65]

Just realised something. Walker's simply letting us solve the puzzle for him ;)

The best kind of slaves are those that think they are doing the work of their own free will :)

that is a good idea. . . :p

No, I am not looking at any of the solutions until after a winner is chosen. If you want I can post my solutions to prove it.

[jjw123]

i did this after seeing your post, and i managed to get from 1 to 52, then got bored and gave up, may post it later on

[Magick_Miner]

I think im at 86/100 at the moment.

Done 1 through 70 consecutively.

Struggling now, will post solutions when im done.




EDIT: Updated progress

[Magick_Miner]

And im done at 86/100

Here they are:

Spoiler

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1= (3-2)*1+0
2= (3*0)+(2*1)
3= (2*0)+(3*1)
4= (3+1)+(2*0)
5= (3+2)*1+0
6= 3+2+1+0
7= (3*2)+1+0
8= (3+2-1+0)!!
9= 12-3+0
10= 3^2+1+0
11= 13-2+0
12= 12+(3*0)
13= 13+(2*0)
14= 13+2^0
15= 13+2+0
16= 20-3-1
17= 20-3*1
18= 21-3+0
19= 20-1^3
20= 21-3^0
21= 21+3*0
22= 23-1+0
23= 23+1*0
24= 23+1+0
25= 20+(3!-1)
26= (3+1)!+2+0
27= 30-2-1
28= 30-2*1
29= 30-2+1
30= 31-2^0
31= 32-1+0
32= 32+1*0
33= 32+1+0
34= 102/3
35= 210/(3!)
36= 3*12+0
37= 3*12+0!
38= (3*2)!!-10
39= 13*(2+0!)
40= (3-1)*20
41= (3!)!!-(2+1)!-0!
42= 32+10
43= (3!)!!-(2+1)!+0!
44= (3+1)!+20
45= (10/2)!!*3
46= (3!)^2+10
47= (2*3)!!-1+0
48= (2*3)!!+(1*0)
49= (2*3)!!+1+0
50= (3+2)*10
51= 31+20
52= (3!)!!+2+1+0!
53= (3!)!!+(2+1)!-0!
54= (3!)!!+(2+1)!+0
55= (3!)!!+(2+1)!+0!
56= (3!)!!+10-2
57= (20-1)*3
58= 3!*10-2
59= 3*20-1
60= 20*3*1
61= 20*3+1
62= 31*2+0
63= 21*3+0
64= 2^(3!)*1+0
65= 130/2
66= 3!*(12-0!)
67= 201/3
68= (3!)!!+20*1
69= (3!)!!+21+0
70= 210/3
71=
72= (2*3)!/10
73=
74= ((3!)!/10)+2
75=
76= ((3!)!!-10)*2
77=
78= 3!*(12+0!)
79=
80= 20*(3+1)
81=
82=
83=
84=
85=
86= ((3!)!!)*2-10
87=
88=
89=
90= 30*(2+1)
91=
92= ((3!)!!-1-0!)*2
93= ((3!)!!-1)*2-0!
94= ((3!)!!)*2-1-0!
95= ((3!)!!)*2-1+0
96= ((3!)!!)*2*1+0
97= 10^2-3
98= ((3!)!!)*2+1+0!
99= ((3!)!!+1)*2+0!
100= ((3!)!!+1+0!)*2

That was very interesting, thanks for posting.

[Ceunon]

Heh, have you guys ever heard of the The Four Fours?

Pretty similar to this one. You can get from 1 to 100 using, well, 4 4s (you must always use the 4 of them).