hint:

**Spoiler:**

**Moderators:** gmalivuk, Moderators General, Prelates

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:**

hint:

Okay.

1-12

**Spoiler:**

1-12

the tree wrote:4 = 2!

You might want to rethink this one...

G4!!

Grob FTW,

Hello. Smithers. You're. Quite good. At. Turning. Me. On.

Grob FTW,

Hello. Smithers. You're. Quite good. At. Turning. Me. On.

- skeptical scientist
- closed-minded spiritualist
**Posts:**6142**Joined:**Tue Nov 28, 2006 6:09 am UTC**Location:**San Francisco

Continuing where tree left off, same conventions:

**Spoiler:**

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:**

Additionally, for the following numbers I have solutions which I find more aesthetically pleasing than tree's (using only simple functions, no repeating decimals).

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

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:

With two 9’s:

Also it is fairly obvious that

With three 9’s:

Last edited by Qaanol on Thu Mar 10, 2011 3:11 am UTC, edited 6 times in total.

wee free kings

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:**

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

**Spoiler:**

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?

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?

wee free kings

- phlip
- Restorer of Worlds
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**Contact:**

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:

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

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

- skeptical scientist
- closed-minded spiritualist
**Posts:**6142**Joined:**Tue Nov 28, 2006 6:09 am UTC**Location:**San Francisco

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

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.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

If you were allowed to use log, you could get all positive integers:

**Spoiler:**

++$_ wrote:If you were allowed to use log, you could get all positive integers:Spoiler:

If you're allowed to use S (the successor function) you can get all positive integers:

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

letterX wrote:++$_ wrote:If you were allowed to use log, you could get all positive integers:Spoiler:

If you're allowed to use S (the successor function) you can get all positive integers:Spoiler:

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.

wee free kings

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