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$$6~~~~6~~~~6~~~~6 = 1 \Rightarrow \frac{6}{6} \cdot \frac{6}{6} = 1$$ $$6~~~~6~~~~6~~~~6 = 2 \Rightarrow \frac{6}{6} + \frac{6}{6} = 2$$ $$6~~~~6~~~~6~~~~6 = 3 \Rightarrow\frac{6+6+6}{6} = 3$$ $$ \cdots$$ $$6~~~~6~~~~6~~~~6 = 10 \Rightarrow \frac{6!/6}{6 + 6} = 10$$ I can make four $6$ s equal from $1$ to $10$ except $9$.

Please use any math operators but not by putting two number together like $77$. Could it be done? Thanks.

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closed as unclear what you're asking by Von Neumann, Frank, Jean Marie, hardmath, Carl Mummert Feb 10 '17 at 1:40

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You just asked an identical and similarly unclear question. $\endgroup$ – The Count Feb 9 '17 at 22:22
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    $\begingroup$ What a great way to teach new comers. Just down vote their questions. $\endgroup$ – Zaid Alyafeai Feb 9 '17 at 22:24
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    $\begingroup$ @ZaidAlyafeai Not if they repeatedly post low quality questions that are extremely similar, show no sign of improvements, and very likely learn little from past experiences. The first question or so? Sure, don't rain down the downvotes, but there comes a limit. $\endgroup$ – Simply Beautiful Art Feb 9 '17 at 22:26
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    $\begingroup$ @ZaidAlyafeai Luckily, regardless of how new I was, I discovered a thing called copy and paste (and frankly used that for many months). IMHO, if one does not downvote indiscriminately of whether or not a user is new or not, this site will slowly degrade in quality, though that is an argument that has already been taken to meta, if you are interested. My personal advice is downvote while the question is worthy of downvotes, explain why, and then remove the downvotes once the OP has cleaned it up. No harm done. $\endgroup$ – Simply Beautiful Art Feb 9 '17 at 22:38
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    $\begingroup$ See here: i.stack.imgur.com/zUZSM.png $\endgroup$ – Simply Beautiful Art Feb 9 '17 at 23:05
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Assuming we're allowed to use the decimal point (which is often allowed in these kinds of puzzles), we have the following: $$\frac{6}{.6}-\frac{6}{6} = 10-1=9.$$ Note that there's nothing special about the number $6$ here: using the decimal point and the usual operations, you can make $9$ starting with four copies of any number you like.

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    $\begingroup$ This isn't really an answer, but it's fun so I'll leave it as a comment. Let $\mathcal{F}=\{1,2,6,24,\ldots\}$ be the set of factorials of positive integers and let $¡\,\cdot\,\colon\mathcal{F}\to\mathbb{N}$ be the anti-factorial, given by $¡(n!)=n$ (note: since $0\not\in\mathbb{N}$ we are forced to have $¡1=1,$ so this is well-defined). Note that $¡6=3.$ Now we can go wild! For example, $$¡6\times¡6\times(6/6)=9,$$ or better yet, $$\sqrt{(¡6)\times(¡6)\times(¡6)\times(¡6)}=\sqrt{3^{4}}=9.$$ $\endgroup$ – Will R Feb 10 '17 at 1:35
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Let me preface this answer by saying it depends on an abuse of notation, so may not be optimal. That said, it's still a neat little trick!

Following the famous idea of Dirac, I'll give a solution that uses $\log$ and $\sqrt[6]{\cdot}$, but can be used to get any number.

Consider the expression: $- \log_6 \log_6 \sqrt[6]6 = -\log_6 1/6 = 1$

Now to achieve the desired result, note that if we keep applying the root function (and this is where we will have to abuse notation to avoid continuously writing sixes), we can take higher roots, and every time we apply a root inside, the output increases by $1$. So the solution to the OP is the special case where we apply the root function 9 times.

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  • $\begingroup$ is $\log$ a operator ? $\endgroup$ – A---B Feb 9 '17 at 23:07
  • $\begingroup$ Well, when Dirac worked on the same problem with $4$'s, he seemed to think so. Of course, you can try to play the game without these operations - I just like this particular solution because it lets you express any number. $\endgroup$ – Alfred Yerger Feb 9 '17 at 23:10
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The greatest common divisor (gcd) of sum from $6$ to $12$ and $6$ factorial is equal to $9$.

$GCD(\sum_{6}^{6+6},6!)=9$

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Oh, I am late, @will already posted a similar answer.

$$ \lceil|(6\times .6) - 6|\rceil + 6$$

$\lceil\cdot\rceil$ is the ceil function.

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  • $\begingroup$ @hardmath done. $\endgroup$ – A---B Feb 10 '17 at 8:53

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