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Chuck · Daedalian Member

93 = 2^7 - 35

Athene · Guest

94=21+73

Chuck · Daedalian Member

95 = (17+2) × 5

Chuck · Daedalian Member

96 = 75 + 21

Chuck · Daedalian Member

97 = 2^7 - 31

Chuck · Daedalian Member

98 = 75 + 23

GH · Daedalian Member

99 = 2 x 53 - 7
100 = 2 x 53 - 7 + 1

CrystyB · Misunderstood Guy

How 'bout these? No use of '%', and any use of log (which is binary-log) can be worked-around. PLUS I NEVER TANGLE WITH THE BLOCKS! :P

0 = 2x3-(1+5)=1+2-3=2+3-5=2+5-7
1=1
2=2
3=3
4=1+3
5=5
6=1+5
7=7
8=1+7
9=1+3x5-7
10=1+2x7-5
11=1+2x5
12=1+2x7-3
13=3+2x5=1+3x(5-log2)
14=1+3x5-2
15=1+2x7=1+3x5-log2
16=1+3x5
17=1+3x7-5
18=2+3x7-5
19=5+2x7=1+3x(7-log2)
20=1+3x7-2
21=1+5x(7-3)=1+3x7-log2
22=1+3x7
23=2+3x7
24=3+7x(5-2)
25=5+3x7-1
26=1+5x(7-2)
27=3x(2+7)
28=3+5x(7-2)
29=1+7x(5-log2) = 31-2
30=2+7x(5-1)
31=3+7x(5-1)=1+5x(7-log2)
32=2+5x(7-1)
33=1+5x7-3
34=1+5x7-2
35=5x7=1+5x7-log2
36=1+5x7
37=2+5x7
38=3+5x7
39=7x(1+5)-3=5x(1+7)-log2
40=7x(1+5)-2
41=7x(1+5)-log2 = 35+7-1
42=5x(2+7)-3
43 = 75-32
44=5x(2+7)-1
45=5x(2+7)
46=7x(2+5)-3
47 = 5x7+12
48=5x(3+7)-2
49=5x(3+7)-1
50=5x(3+7)
51 = 51
52 = 52
53 = 53
54=7x(3+5)-2
55=7x(3+5)-1
56=7x(3+5)
57 = 57
58 = 57+1
59 = 57+2
60 = 57+3
61 = 57+2x(3-1)
62 = 57+2x3-1
63 = 57+2x3
64 = 75-13+2
65 = 12+53
66 = 75-12+3
67 = 53+2x7
68 = 57+13-2
69 = 57+12
70 = 57+13
71 = 75-(1+3)
72 = 75-3
73 = 75-2
74 = 75-1
75 = 75
76 = 75+1
77 = 75+2
78 = 75+3
79 = 53+27-1=75+2x(3-1)
80 = 75+2x3-1=53+27
81 = 75+2x3=27x3
82 = 12x7-5+3=25x3+7
83 = 17x5-2
84 = 12x7
85 = 17x5=137-52
86 = 12x7+5-3
87 = 12x7+3
88 = 17x5+3
89 = 12x7+5
90 = (21-3)x5
91 = 7x13
92 = 25x3+17
93 = 7x13+2
94 = 7x13+5-2
95 = 5x(17+2)
96 = 7x13+5=27x3+15
97 = 75+23-1
98 = 23x5-17
99 = 51x2-3
100 = 15x7-(2+3)=25x(3+1)

robichelli · MI:6 Agent

That's very nice, but because the numbers are made of blocks, they cannot be used more than once. You've twisted the reasoning out of the lovely puzzle Jingle47 produced.

Zealot · Daedalian Member

He didn't use them more than once. He just provided alternate solutions.

cubestudent · 3D Member

in a similar vein, one of my teacher once asked us back in. . . 1996? argh! i'm getting way too old!

anyway, told us to use the digits: 1, 9, 9, and 6, in that order, using whatever operations we wished, to get all the numbers between 1 and 1996. it was an extra credit problem, which meant it didn't violate my normal "I will not do any homework" rule, so I made it a fair portion of the way through the numbers.

since the blocks one recieved so much interest, perhaps we could do the first hundred or so here?

in general, i would say the more 'conventional' or 'simple' the operation the better.

1 = 1^996

robichelli · MI:6 Agent

Very true. They are alternate, yet adequade solutions to the puzzle, and I am sorry if I have offended CrystyB (Misunderstood Guy) in any way shape, form, or manner.
(Edited to prove that I cannot spel)

[This message has been edited by robichelli (edited 04-02-2002 07:18 PM).]

CrystyB · Misunderstood Guy

Lovely problem, 1996 one. And i won't be able to solve it by computer either

...

No robi, i am not offended that easily (though mispelling my name would be quicker to offend me, even that is not enough

- hehe, TY 4 the edit, Crystal is way too feminine for me... ).

PS 2=-(1^9)+9-6
3=(1^9)*(9-6)
4=(1^9)+9-6
5=(-1^99)+6
6=(1^99)*6
7=(1^99)+6
8=1+(9/9)+6
...
15=(1+9)*9/6

[This message has been edited by CrystyB (edited 04-02-2002 07:17 PM).]

Athene · Guest

2=Ö(19-9-6)
3=(1^9)*(9-6)
4=19-9-6
5=-1+9-9+6
6=1^(9-9)*6
7=1/(9-9)+6

Chuck · Daedalian Member

*** Divide by zero error in line 7 ***

[This message has been edited by Chuck (edited 04-02-2002 06:50 PM).]

CrystyB · Misunderstood Guy

9=(-1*(Ö9))*((Ö9)-6)
10=1^9+(Ö9)+6
11=(-1+(Ö9))+(Ö9)+6

LOL Chuck!

[This message has been edited by CrystyB (edited 04-02-2002 06:52 PM).]

Athene · Guest

7=1/(9/9)+6

CrystyB · Misunderstood Guy

12=1*(Ö9)+(Ö9)+6
13=1+(Ö9)+(Ö9)+6

i'll let y'all take over. I just wanna do my birthday, but i'm out of ideas. I'll think of it till tomorrow, BUT i wonder. What exactly can we do? I don't know too many operations, so what am i allowed to bring in?

528=?
-199+6!=521, bah SO CLOSE!!!

At least i got WWI starting year: 1939=1+9!!+9!!+6!!

(the double factorial is x*(x-2)*(x-4)*...*y, where y=1 if x odd or y=2 if x even. EG 10!!=3840, 8!!=384, 6!!=48 ; 9!!=945...)

[This message has been edited by CrystyB (edited 04-02-2002 07:33 PM).]

Zealot · Daedalian Member

I can get 523...

cubestudent · 3D Member

you can use any operation you want, though for most cases I'd prefer if you stuck to common operations.

[This message has been edited by cubestudent (edited 04-02-2002 07:30 PM).]

CrystyB · Misunderstood Guy

So you're not going to point out the obvious ones i am missing?

And how in the world would you get 1980 through COMMON operations???

[This message has been edited by CrystyB (edited 04-02-2002 07:37 PM).]

Bicho the Inhaler · Daedalian Member

16 = 1^9 + 9 + 6

Zealot · Daedalian Member

14 = -1^9 + 9 + 6

Bicho the Inhaler · Daedalian Member

...or 16 19 + Ö9 - 6

17 = -1^9 + (Ö9) * 6
18 = 1^9 * (Ö9) * 6
19 = 1^9 + (Ö9) * 6

edit: dumb mistake

[This message has been edited by Bicho the Inhaler (edited 04-02-2002 07:55 PM).]

Zealot · Daedalian Member

never mind.

[This message has been edited by Zealot (edited 04-02-2002 07:56 PM).]

Bicho the Inhaler · Daedalian Member

20 = -1 + 9 * (Ö9) - 6

Bicho the Inhaler · Daedalian Member

21 = (1 + Ö9)! + (Ö9) - 6

Zealot · Daedalian Member

21 = 1 * 9 * Ö9 - 6
22 = 1 + 9 * (Ö9) - 6

Zealot · Daedalian Member

23 = -1 + 9 + 9 + 6
24 = (-1 + 9) * (9 - 6)
25 = 1 + 9 + 9 + 6
26 = -1 + 9^(9/6)
27 = 1 * 9^(9/6)
28 = 1 + 9^(9/6)

[e]I'm an idiot too. 9^(9 - 6) indeed...[/e]

[This message has been edited by Zealot (edited 04-02-2002 08:10 PM).]

Zealot · Daedalian Member

Sorry, I had typoed and put 27 when I meant to put 26.

Bicho the Inhaler · Daedalian Member

You forgot 26:

26 = -1 + 9^(9/6)
edit: UBB

[This message has been edited by Bicho the Inhaler (edited 04-02-2002 08:08 PM).]

Bicho the Inhaler · Daedalian Member

wait...for 26-28, you either need to replace "9-6" with "9/6" or "9^" with "9*".

Zealot · Daedalian Member

Yeah, I saw that after I read your post

Bicho the Inhaler · Daedalian Member

29 = -1 - (Ö9)! + (Ö9)! * 6

Bicho the Inhaler · Daedalian Member

30 = -1 * (Ö9)! + (Ö9)! * 6

Bicho the Inhaler · Daedalian Member

31 = 1 - (Ö9)! + (Ö9)! * 6
32 = -1 - (Ö9) + (Ö9)! * 6
33 = -1 * (Ö9) + (Ö9)! * 6

CrystyB · Misunderstood Guy

34 = 1 - (Ö9) + (Ö9)! * 6
35 = -1^9 + (Ö9)! * 6
36 = 1^9 * (Ö9)! * 6

Bicho the Inhaler · Daedalian Member

34 = 1 + (Ö9)! + (Ö9) * 6

Zealot · Daedalian Member

If you have the last post in the thread it's easier to just edit it... Unless you're shamelessly boosting post count

Bicho the Inhaler · Daedalian Member

True...but I think it's only fair that if someone beats me to it (quite likely in threads like this), his/her post appears before mine. (Overruled

)