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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.

projectilemotion
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Gilbert Chiang
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  • Try to explain what you did – user326159 Feb 09 '17 at 22:21
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    You just asked an identical and similarly unclear question. – The Count Feb 09 '17 at 22:22
  • See my edits on your four 7s question to make this readable. Note that two spaces at the end of a typed line will produce a line break on display. And also note the tags I switched to. – Joffan Feb 09 '17 at 22:23
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    What a great way to teach new comers. Just down vote their questions. – Zaid Alyafeai Feb 09 '17 at 22:24
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    @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. – Simply Beautiful Art Feb 09 '17 at 22:26
  • @SimplyBeautifulArt, It is difficult for new comers to know the rules here and some people don't give them a chance. This is the second question of the OP. It takes months to post a good question. – Zaid Alyafeai Feb 09 '17 at 22:34
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    What operations are you allowed to use? – Tim Thayer Feb 09 '17 at 22:37
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    @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. – Simply Beautiful Art Feb 09 '17 at 22:38
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    See here: https://i.stack.imgur.com/zUZSM.png – Simply Beautiful Art Feb 09 '17 at 23:05
  • Possible duplicate of How to make four 7 s equal to 4 and to 10? – Carl Mummert Feb 10 '17 at 01:40
  • @ZaidAlyafeai I usually agree with you, but this person had rep, knew what they were doing, and had just been through the same circus with another question. Unacceptable. – The Count Feb 10 '17 at 02:08
  • if $\sqrt{\cdot}$ is allowed, then $$\frac{6\times 6}{\lceil \sqrt{6+6} \rceil } = \dfrac{36}{4} = 9,$$ where $\lceil \cdot \rceil$ is round-up. – Oleg567 Feb 13 '17 at 15:25

4 Answers4

6

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.

Will R
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    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, $$\sqrt{(¡6)\times(¡6)\times(¡6)\times(¡6)}=\sqrt{3^{4}}=9.$$ – Will R Feb 10 '17 at 01:35
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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.

A. Thomas Yerger
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  • is $\log$ a operator ? –  Feb 09 '17 at 23:07
  • 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. – A. Thomas Yerger Feb 09 '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 $9$.

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

LAAE
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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.