Four nines puzzle

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sinal
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Four nines puzzle

Postby sinal » Thu Mar 10, 2011 12:07 am UTC

This is similar to the older four fours puzzle. Using only four nines and operations, you have to get as many whole numbers as possible. You have to use four nines, can't use three. For example, 99-99 would be zero. You can't use any functions with letters, like log,sin,cos, ect. Other than that you can use any mathematical operation or function, ceiling and floor functions included

hint:
Spoiler:
use .9(bar) for 1. It helps a lot.

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Re: Four nines puzzle

Postby the tree » Thu Mar 10, 2011 12:48 am UTC

Okay.
1-12
Spoiler:
Assume 1 is to be read as .9(bar)

1 = 99/99
2 = 9/9 + 9/9
3 = (9 + 9 + 9)/9
4 = 2!
5 = 9.9(bar)/(1 + 1))
6 = 3!
7 = (9 - 1 -1)/1
8 = (9 - 1) * 1 * 1
9 = 9 * 1 * 1 * 1
10 = 9.9(bar)*1*1
11 = (9.9(bar)+1)*1
12 = 9.9(bar)+1+1

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Re: Four nines puzzle

Postby OverBored » Thu Mar 10, 2011 1:11 am UTC

the tree wrote:4 = 2!


You might want to rethink this one...
Spoiler:
1+1+1+1 works fine though
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Re: Four nines puzzle

Postby skeptical scientist » Thu Mar 10, 2011 1:25 am UTC

Continuing where tree left off, same conventions:
Spoiler:
13: 9/.9+9/√9
14: 9+9/(.9+.9)
15: √9*9/(.9+.9)
16: 9+9-1-1
17: 9+9-9/9
18: 9+9*9/9
19: 9+9+9/9
20: 99/9+9
21: 9+9+9/√9
22: 9+9+√9+1
23: 9√9-√9-1
24: 9+9+√9+√9
25: 9√9-√9+1
26: 9+9+9-1
27: 9+9+√(9*9)
28: 9+9+9+1


Additionally, for the following numbers I have solutions which I find more aesthetically pleasing than tree's (using only simple functions, no repeating decimals).
Spoiler:
4: 9-9/(.9+.9)
7: 9-(9+9)/9
8: (9*9-9)/9
9: 9+(9-9)/9
10: 99/9.9
12: (99+9)/9
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Re: Four nines puzzle

Postby Qaanol » Thu Mar 10, 2011 1:35 am UTC

Spoiler:
21 = (9 + 9 + √9) × 1
22 = 9 + 9 + √9 + 1
23 = (9 - 1) × √9 - 1
24 = (9 - 9/9) × √9
25 = (9 - 1) × √9 + 1
26 = 9 × √9 - 9/9
27 = 9 × √9 + 9 - 9
28 = 9 × √9 + 9/9
29 = (9 + 1) × √9 - 1
30 = (9 + 9/9) × √9


I think it might be helpful for notation purposes to list numbers that can be obtained using exactly 1, 2, or 3 numeral 9’s, so we can use those the way we are using 1 currently.

With one 9:
Spoiler:
1 = .9(bar)
3 = √9
6 = (√9)!
9 = 9
720 = ((√9)!)!
362,880 = 9!

(multiple factorials, etc.)


With two 9’s:
Spoiler:
0 = 9 - 9
1 = 9 / 9
2 = .9(bar) + .9(bar)
3 = √9 × .9(bar)
4 = √9 + .9(bar)
5 = (√9)! - .9(bar)
6 = 9 - √9
7 = (√9)! + .9(bar)
8 = 9 - .9(bar)
9 = 9 × .9(bar)
10 = 9 + .9(bar)

12 = 9 + √9
15 = 9 + (√9)!
18 = 9 + 9
24 = (√9 + .9(bar))!
27 = 9 × √9
36 = (√9)! × (√9)!
54 = 9 × (√9)!
57 = [imath]\left\langle\begin{array}{c}(√9)!\\ .\overline{9}\end{array}\right\rangle[/imath]
80 = ((√9)!)! / 9
81 = 9 × 9
84 = [imath]{9\choose\sqrt{9}}[/imath]
99 = 99
120 = ((√9)!)! / (√9)!
216 = (√9)! ^ √9
240 = ((√9)!)! / √9
302 = [imath]\left\langle\begin{array}{c}(√9)!\\ √9\end{array}\right\rangle[/imath]
502 = [imath]\left\langle\begin{array}{c}9\\ .\overline{9}\end{array}\right\rangle[/imath]
504 = 9! / ((√9)!)!
648 = ((√9)!)! × .9
711 = ((√9)!)! - 9
714 = ((√9)!)! - (√9)!
717 = ((√9)!)! - √9
719 = ((√9)!)! - .9(bar)
720 = ((√9)!)! × .9(bar)
721 = ((√9)!)! + .9(bar)
723 = ((√9)!)! + √9
729 = ((√9)!)! + 9
800 = ((√9)!)! / .9
1440 = ((√9)!)! + ((√9)!)!
2160 = ((√9)!)! × √9
4320 = ((√9)!)! × (√9)!
5040 = ((√9)! + .9(bar))!
6480 = ((√9)!)! × 9
19,683 = √9 ^ 9
40,320 = 9! / 9
46,656 = (√9)! ^ (√9)!
60,840 = 9! / (√9)!
88,234 = [imath]\left\langle\begin{array}{c}9\\ √9\end{array}\right\rangle[/imath]
120,960 = 9! / √9

(etc.)


Also it is fairly obvious that
Spoiler:
Anything you can get with N nines you can get with N+1 nines, by multiplying by .9(bar)


With three 9’s:
Spoiler:
1-21: possible

23-28: possible

30: possible

32-33: possible

35-37: possible

(etc.)
Last edited by Qaanol on Thu Mar 10, 2011 3:11 am UTC, edited 6 times in total.
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Re: Four nines puzzle

Postby Dopefish » Thu Mar 10, 2011 2:19 am UTC

In similar problems I've had objections saying that [imath]\sqrt{}[/imath] is implicitly [imath]\sqrt[2]{}[/imath], and so not allowed due to the use of a 2. Are plain square root's fair game here?

Spoiler:
(Also, I'd probably want to avoid .9 bar as much as possible since having a 1 to work with almost makes things too easy. :P )

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Re: Four nines puzzle

Postby Qaanol » Thu Mar 10, 2011 2:40 am UTC

Hmm. Well I’d like to suggest we either disallow √ or we disallow floor and ceiling.

Spoiler:
Otherwise they, along with factorials, can be nested indefinitely with just one 9 to get (I suspect) any integer.


Which leads to my first side challenge:

Using the rules as originally stated (so all of the above are fair game) how can you get 11 with just one 9?
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Re: Four nines puzzle

Postby phlip » Thu Mar 10, 2011 3:33 am UTC

Qaanol wrote:Which leads to my first side challenge:

Using the rules as originally stated (so all of the above are fair game) how can you get 11 with just one 9?

Some experimenting reveals:
Spoiler:
[math]\left \lceil \sqrt{\left \lfloor \sqrt{\sqrt{\sqrt{9}!!}} \right \rfloor!} \right \rceil[/math]

Working back from the answer helped... 112 is very close to 5!, so I was aiming for [imath]\left \lceil \sqrt{5!} \right \rceil[/imath].

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Re: Four nines puzzle

Postby the tree » Thu Mar 10, 2011 11:43 am UTC

OverBored wrote:
the tree wrote:4 = 2!
Urm... it was one in the morning? Does that count as an excuse?

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Re: Four nines puzzle

Postby skeptical scientist » Thu Mar 10, 2011 1:15 pm UTC

Qaanol wrote:Hmm. Well I’d like to suggest we either disallow √ or we disallow floor and ceiling.

Spoiler:
Otherwise they, along with factorials, can be nested indefinitely with just one 9 to get (I suspect) any integer.

I suspect you are correct. So far all I've been able to show is that for all n there are m and k such that \(|n-\sqrt[2^k]{m!}|<1\). If we could show that m<n, we could use induction to get all natural numbers representable with one 9, but I haven't gotten that far.

Spoiler:
To get n, look at the sequence of intervals [n2,(n+1)2], [n4, (n+1)4], [n8, (n+1)8]... We want to show that one of these intervals contains a factorial. The right endpoint of the interval with left endpoint nk is ((n+1)/n)k times the left endpoint. Factorial growth is superexponential, so for sufficiently large k, if m>((n+1)/n)k, then m!>nk. So for sufficiently large k, if m!≤nk, then (m+1)≤(n+1)k, which means there is a factorial in one of these intervals (in fact, all but finitely many of the intervals contain a factorial).

It's possible that a good estimate will show that "sufficiently large k" actually gives m<n whenever n>c, which means that if all numbers ≤c are representable with a single 9, then all integers are.
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Re: Four nines puzzle

Postby Qaanol » Thu Mar 10, 2011 4:13 pm UTC

Clever approach. Don’t forget

Spoiler:
The ‘!’ notation is also used for the ∏ function, so [imath]x! = \Pi(x) = \Gamma(x+1) = x\Gamma(x)[/imath] meaning you can take the factorial of non-integers.
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Re: Four nines puzzle

Postby ++$_ » Fri Mar 11, 2011 12:16 am UTC

If you were allowed to use log, you could get all positive integers:
Spoiler:
[math]n = -\log_{\sqrt{9}!/\sqrt{9}}\left(\log_\sqrt{9} \mbox{sqrt}^{n+1}(9)\right),[/math]where the superscript indicates function composition.

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Re: Four nines puzzle

Postby letterX » Fri Mar 11, 2011 12:52 am UTC

++$_ wrote:If you were allowed to use log, you could get all positive integers:
Spoiler:
[math]n = -\log_{\sqrt{9}!/\sqrt{9}}\left(\log_\sqrt{9} \mbox{sqrt}^{n+1}(9)\right),[/math]where the superscript indicates function composition.


If you're allowed to use S (the successor function) you can get all positive integers:
Spoiler:
[math]n = S^n(99-99)[/math]

Having circuit depth linear in the number you're creating is really uninteresting.

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Re: Four nines puzzle

Postby Qaanol » Fri Mar 11, 2011 10:41 pm UTC

letterX wrote:
++$_ wrote:If you were allowed to use log, you could get all positive integers:
Spoiler:
[math]n = -\log_{\sqrt{9}!/\sqrt{9}}\left(\log_\sqrt{9} \mbox{sqrt}^{n+1}(9)\right),[/math]where the superscript indicates function composition.


If you're allowed to use S (the successor function) you can get all positive integers:
Spoiler:
[math]n = S^n(99-99)[/math]

Having circuit depth linear in the number you're creating is really uninteresting.

We can probably coopt the ++ notation to get that, right?

((((99-99)++)++)++)++ etc.
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