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Possibly not a Math problem in the true sense of the word, but here goes.

Three Nines - and various math operators = can be turned into various simple numbers.

(9-9)*9 = 0

(9/9)^9 = 1

(9+9)/9 = 2

sqrt (9+9-9) = 3

(sqrt(9) +9/9) = 4

(sqrt(9!) - 9/9 = 5

sqrt (9*9) - sqrt (9) = 6 (or 9 - (9/Sqrt (9) )

9 - sqrt 9 + .9... = 7

9 - (9/9) = 8

9 * (9/9) = 9

9 +(9/9) = 10

99/9 = 11

9 + (9/sqrt(9)) = 12

anything for 13 and beyond?

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This is a bit off topic. (Typo 10 not 20) –  AD. Nov 6 '10 at 13:51
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A related question: how many distinct numbers can be made by combining three 9s using the operations +,-,*,/,^,and ! –  Seamus Nov 6 '10 at 14:56
    
Oh wait, infinitely many, because you can just stack factorials to make bigger and bigger numbers. So leave off !... –  Seamus Nov 6 '10 at 14:57
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I'm a fan of $9^{9^9}$ myself... :) Quick, what are the first and last digits? –  J. M. Nov 6 '10 at 15:13
    
@J.M. I make it 4 and 9 –  Ross Millikan Nov 11 '10 at 16:54
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closed as off topic by Qiaochu Yuan Sep 1 '11 at 5:49

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5 Answers

$7=(\sqrt{9})!+\frac{9}{9}$. Also for 5 you need to move the ! outside the parens.
$18=\sqrt{(\sqrt{9})!!+9}-9=\sqrt{9}^{\sqrt{9}}-9$
and you can sqrt the last 9 and add and subtract to get a few more

$13=\frac{9}{.9}+\sqrt{9}$
If you allow the overline for a repeating decimal or the floor function (floor(sqrt(9)) you can convert a 9 into a 1. This gives new flexibility

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As Derek points out, we can blow the question wide open by allowing operations not included on basic calculators. In fact, even a single nine could be quite productive if we allow (say) floor, ceiling, factorial, and the natural logarithm:

$$\lfloor\ln\ln9\rfloor=0$$

$$\lceil\ln\ln9\rceil=1$$

$$\lfloor\ln 9\rfloor=2$$

$$\lceil\ln 9\rceil=3$$

$$\lceil\ln\ln\lceil\ln(9!)\rceil!\rceil=4$$

$$\lfloor\ln\ln\lceil\ln\lceil\ln\lceil\ln(9!)\rceil!\rceil!\rceil!\rfloor = 5$$

$$\lceil\ln 9\rceil!=6$$

$$\lceil\ln\lceil\ln9\rceil!!\rceil= 7$$

$$\vdots$$

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"Representing Numbers Using Only One 4", Donald Knuth, (Mathematics Magazine, Vol. 37, Nov/Dec 1964, pp.308-310). Knuth shows how (using a computer program he wrote) all integers from 1 through 207 may be represented with only one 4, varying numbers of square roots, varying numbers of factorials, and the floor function. It is at jstor.org/pss/2689238, but I don't have access. –  Ross Millikan Nov 7 '10 at 15:12
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Here's a quick cheat for 7.

$$7 = 9 - 9^0 - 9^0.$$

EDIT:

Depending on what functions one's allowed they are easy to come by. Here are some more examples for $7.$

$$ d(9)! + \log_9 9 = \lfloor \sqrt {\sigma(9)} \rfloor ! + \frac{9}{9}$$ $$= \lfloor \log \sqrt{9!} \rfloor + \frac{9}{9} = \lfloor \sqrt{ \log 9^9 } \rfloor + \sqrt{9} = 7.$$

As usual, $d(n)$ is the number of divisors of $n$ and $\sigma(n)$ is the sum of the divisors of $n.$

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Assuming, if we permit sqrt then we also permit raising to other powers. –  Derek Jennings Nov 6 '10 at 13:46
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Usually in these problems, sqrt is permitted because it doesn't involve other numbers. You can raise to powers only if you have the digit available. So 9^9 or 9^sqrt(9) would be OK, but use two of your nines –  Ross Millikan Nov 6 '10 at 14:34
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But that would be 3 nines and 2 zeros. –  Peter Nov 6 '10 at 14:47
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@Ross Millikan; @Peter; You are quite right, which is why I qualified my answer with the word "cheat." However, the rules of the game were not 100% clear aka "various math operators," and I viewed sqrt as including an implicit 1 and 2, or even a 0 and a 5. Anyway, well done Ross for your 7. It's got my upvote! –  Derek Jennings Nov 6 '10 at 16:34
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I recommend ries which finds solutions to this problem, related problems with different digits and operators, and in general finds operations that end up with a specified number.

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$16= 9/.9 +(\sqrt9)!$

$19= 9+ 9/.9$

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