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        riddles >> medium >> Equations from 1998
        
(Message started by: jeremiahsmith on Aug 8^th, 2002, 6:21pm)

Title: Equations from 1998
Post by jeremiahsmith on Aug 8^th, 2002, 6:21pm

I put this puzzle in medium because it's not easy, yet I don't think it's difficult enough for hard.

This is based on an extra credit assignment my brother once had, and I helped him with. To celebrate the new year in 1998, his math teacher gave him this puzzle. There are lot of acceptable operations for this, since it's for early high schoolers.

Using the digits 1, 9, 9, 8, in that order, create equations that equal every number from 0 to 100. (I believe the original puzzle didn't include 0, but I'm sticking it in anyway.)
You're allowed to use +, -, *, /, (), ^, and the square root symbol. (You can also use numbers as the radix of the root. Thus, you could do cube roots if you could find a 3...)
You're allowed to concatenate digits, e.g. using 19 + 9 + 8 to get 36. BUT, you have to use the digits...no fair writing "10" as 1(9 - 9) even though 9 - 9 is 0.
The digits must be written in order. You can't do 98 - 91 for seven, as that rearranges the digits. Do fractions as (x / y) instead of the vertical way.

That's about it, I think. Although me and him only managed to get about two-thirds of the numbers, he informed me later that some kid in his class got all 100.

So, either manage to get all the numbers from 0 to 100, or if you can't, say how many numbers from 0 to 100 are possible. (Most of the numbers have several possible equations.) It might be best to do this as a group effort thing...one person posts a few answers, then the next person posts a few that haven't been found yet...I suppose if you had a more elegant or pretty solution to a number that had been solved, you could do that, but let's not degenerate this into elegance one-upmanship. :)

Title: Re: New puzzle: 1998
Post by william wu on Aug 9^th, 2002, 11:01am

Cool! I think I remember doing something like this in junior high once. Eventually the class compiled 0 to 100. The process was expedited when my Dad and I spent one night finishing the whole list, just so I could bring it in the next day and show everyone up :)

I think this process will be easier if we post solutions sequentially, from 0 to 100. Then we won't have to deal with nasty merges, and figuring out what's left. I'll start it off:

0 = 1*(9 - 9)*8
1 = 1^998
2 = (19 - sqrt(9)) / 8

if you have "more elegant" ways to construct those numbers, feel free to post them; just remember to list the next number needed:

Next Digit Needed: 3

Title: Re: New puzzle: 1998
Post by Eric Yeh on Aug 9^th, 2002, 11:59am

Can someone mail me when it's down to the last five? ;)

Title: Re: New puzzle: 1998
Post by anshil on Aug 9^th, 2002, 1:11pm

2 = 19 - 9 - 8
3 = sqrt(1 + 9 - 9 + 8)

Title: Re: New puzzle: 1998
Post by william wu on Aug 9^th, 2002, 3:23pm

3 = 1 - sqrt(9) - sqrt(9) + 8
4 = 1 * (sqrt(9) + 9 - 8 )
5 = 1 + sqrt(9) + 9 - 8
6 = -1 - (9/9) + 8
7 = -1 + 9 - 9 + 8
8 = 1 * ((9-9) + 8 )
9 = 1 + 9 - 9 + 8

yea i'm bored :(

Next Digit: 10

Title: Re: New puzzle: 1998
Post by Jonathan_the_Red on Aug 9^th, 2002, 3:35pm

10 = (1 + 9) * (9 - 8)
11 = 1 + 9 + 9 - 8

Title: Re: New puzzle: 1998
Post by Alfonso on Aug 11^th, 2002, 3:09am

12=(-1+sqr(9))*9-8
13=-1-sqr(9)+9+8
14=-(1*sqr(9))+9+8
15=1-sqr(9)+9+8
16=(1+(9/9))*8
17=1*sqr(9*9)+8
18=19-(9-8)
19=19*(9-8)
20=19+(9-8)

Title: Re: New puzzle: 1998
Post by william wu on Aug 11^th, 2002, 1:42pm

i don't think your answer for 12 works

(-1+sqr(9))*9-8 = ( -1 + 3 ) * 9 - 8 = 18 - 8 = 10

Title: Re: New puzzle: 1998
Post by jeremiahsmith on Aug 11^th, 2002, 2:43pm

Remember, I said that you could use numbers as the radix in the square root symbol. (And remember, it's the actual symbol you're supposed to have in mind, not a function.)

If I illustrate the square root symbol as v^-, you can write 12 as 1 + 9 + (v^-9)v^-8

You take the square root of 9 to get 3, and then use that 3 to get the cube root of 8, which is 2. :D

Title: Re: New puzzle: 1998
Post by Alfonso on Aug 11^th, 2002, 3:31pm

on 08/11/02 at 13:42:13, william wu wrote:

i don't think your answer for 12 works

(-1+sqr(9))*9-8 = ( -1 + 3 ) * 9 - 8 = 18 - 8 = 10

Sorry, I don't know what I was thinking. :-[
Here is another solution: 12 = 1 + sqr(sqr(9*9)) + 8 = 1 + 3 +8

Last comment by Jeremiah is interesting...

Title: Re: New puzzle: 1998
Post by Jonathan_the_Red on Aug 11^th, 2002, 5:32pm

21 = 1 + 9 + sqrt(9) + 8
22 = 1 - sqrt(9) + (sqrt(9) * 8)
23 = -1 + ((9 / sqrt(9)) * 8)
24 = 1 * (9 / sqrt(9)) * 8
25 = 1 + ((9 / sqrt(9)) * 8)
26 = -1 + sqrt(9) + sqrt(9) * 8
27 = 1 + 9 + 9 + 8

Title: Re: New puzzle: 1998
Post by jkemp on Aug 11^th, 2002, 11:15pm

Sorry about the blanks...

28 = 1 + sqrt(9) + sqrt(9) * 8
29 =
30 =
31 =
32 = -1 + 9 + sqrt(9) * 8
33 = 1 * 9 + sqrt(9) * 8
34 = -1 + sqrt(9) * 9 + 8
35 = 1 * sqrt(9) * 9 + 8
36 = 1 + sqrt(9) * 9 + 8
37 =
38 = (1 + 9) * sqrt(9) + 8

Next is 29

Title: Re: New puzzle: 1998
Post by anshil on Aug 11^th, 2002, 11:54pm

Code:

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

(modify: corrected a wrong calculation (93)
(modify: corrected a wrong calculation (31)

Title: Re: New puzzle: 1998
Post by anshil on Aug 12^th, 2002, 12:01am

It seems I can't post my Python script, Its too long and unfortunally BBBoard seems to destroy the proper indention which is critical for python :( If you mail me. riddlespam [at] kittenberger.net, I'll reply with the raw text if you're interested. (you will need at least Python 2 to run it.) Also I don't have a webserver to upload it anywhere.

"n v/~ x" means the n-th root of x.

It runs ~4 - 5 hours on my athlon 700.

for the unresolved entries, either the alg. misses something or they aren't possible with this conditions.

Title: Re: New puzzle: 1998
Post by Alfonso on Aug 13^th, 2002, 1:19am

anshil:
I've been trying to find a solution for 31 but I was unable so finally I gave up and looked at the listed solution, but I just don't understand it. Can you explain it?
31: ((1v/~sqrt(9))9-8) = 31

it seems useless to take the 1-root of 3 (it's cleaner 1*3) and then it illegally concatenates the 3 with 9, Or am I wrong?

Title: Re: New puzzle: 1998
Post by jkemp on Aug 13^th, 2002, 1:23am

Well done Alfonso, looks like you've found the bug in anshil's program. Concatenation of formula results is not allowed, but anhil's program allows it because concatenation of the initial digits is allowed.

Title: Re: New puzzle: 1998
Post by anshil on Aug 13^th, 2002, 2:47am

Oops, yes I had a bug that let the result of nth-root operation beeing traited as a digit. I've corrected it, and well yes it doesn't find a solution now for 31 :(

Title: Re: New puzzle: 1998
Post by Lightboxes on May 30^th, 2003, 11:58am

If you guys have the solutions for the remaining blanks...please post them.

for 37 I got:
[hide]1+sqrt(sqrt( (sqrt(9) + sqrt(9)) ^8 ))[/hide]

Title: Re: New puzzle: 1998
Post by Lightboxes on Jun 2^nd, 2003, 8:59pm

can we do this?:
[hide]1+sqrt(9)*9+sqrt(sqrt(sqrt(sqrt(...(8 ) = 29?

[/hide]edited - I also forgot...also, how about doing this?:
[hide]sqrt(sqrt( (-1+9) + ^ 8 + sqrt(9) )) = 67 BUT putting the "sqrt(9)" neatly and tiny just to the left and under the "^ 8"???

I'm hoping someone can use 9^4.5 (sqrt[9^9]).
Since it comes out as a whole number.
[/hide]

Title: Re: New puzzle: 1998
Post by 57 =Margit Schubert-While on Jul 13^th, 2003, 6:37am

57 = 19*sqrt(sqrt(sqrt(sqrt(9)^8)))

Cab somebody update the list ?

Title: Re: New puzzle: 1998
Post by Margit Schubert-While on Jul 13^th, 2003, 6:40am

57 = 19*sqrt(sqrt(sqrt(sqrt(9)^8)))

Can somebody update the list ?

(Dunno what happened to my previous reply)

Title: Re: New puzzle: 1998
Post by rmsgrey on Jul 14^th, 2003, 4:56am

Here in London, England, there's an inter-school mathematics tournament (teams of 4 from each school). One of the rounds in each match is a team round - where each team collaboratively attempts to find as many solutions as possible to a given puzzle. One that comes up every year is creating as many numbers in a given time as possible in a given range using the appropriate year...

The last year I took part was 1997, so I never tackled 1998 myself, but somewhere at home, I probably still have partial lists for 1994 onwards.

Title: Re: New puzzle: 1998
Post by aero_guy on Jul 22^nd, 2003, 7:27am

It is odd that you allow sqrt when this is the same as 2v/~ you may as well allow squaring of things.

Title: Re: New puzzle: 1998
Post by Jeremiah Smith on Jul 23^rd, 2003, 12:35pm

Exponentials ARE allowed... the reason square roots are allowed is because the radical symbol is allowed, and using the radical symbol with no radix is a square root. There's no such convention for exponentiation.