# Use math symbols to produce 6 [duplicate]

How can we get (6) from these numbers?

$2$ $2$ $2 = 6$

$3$ $3$ $3 = 6$

$4$ $4$ $4 = 6$

$5$ $5$ $5 = 6$

$6$ $6$ $6 = 6$

$7$ $7$ $7 = 6$

$8$ $8$ $8 = 6$

$9$ $9$ $9 = 6$

$12$ $12$ $12 = 6$

$15$ $15$ $15 = 6$

• Please specify — which operations can we use? Can we use the square root function $\sqrt{n}$ ? Can we use the floor/ceiling functions to round (like I have in my answer)? Sep 7, 2018 at 20:21
• @JonMarkPerry it's worth a discussion whether this is a dupelicate. In a question like this, one person might ask for 0 through 9, and then another person might ask for 0 through 15. Is the whole question a duplicate if part of the question is a duplicate? Sep 7, 2018 at 22:20
• if you want you could carry the concept on indefinitely, but the principle is the same as in the duplicate. @Hugh
– JMP
Sep 7, 2018 at 22:30
• @benasfan is my answer better now? Sep 8, 2018 at 17:57
• @Hugh much much better for 12, still that for 15 doesn't require complex work. Believe me it is as easy as the whole question is. An answer with a simple calc and a pencil would be more than sufficient like that of 12. Sep 8, 2018 at 18:45

I made some searches on Google and found:

(Click on equation for WolframAlpha link)
2. $2 + 2 + 2 = 6$
3. $3 \times 3 - 3 = 6$
4. $\sqrt{4} + \sqrt{4} + \sqrt{4} = 6$
5. $5 + \frac{5}{5} = 6$
6. $6×\frac{6}{6} = 6$
7. $7 - \frac{7}{7} = 6$
8. $8 - \sqrt{\sqrt{8 + 8}} = 6$
9. $\sqrt{9 \times 9} - \sqrt{9} = 6$
12. $\sqrt{12 + 12 + 12} = 6$ or $12 \log_{12} {\sqrt{12}}$ where $\log_xy$ is the base $x$ logarithm.
15. $\lfloor\sqrt{\sqrt{15 \times 15}}\rfloor + \lfloor\sqrt{15}\rfloor = 6$ where $\lfloor x \rfloor$ is the mathematical floor function, which "takes as input a real number and gives as output the greatest integer less than or equal to the input number" or $\pi(15 + \frac{15}{15}) = 6$ where $\pi(x)$ is the prime counting function or $\Gamma(\sqrt{15 + \frac{15}{15}})$ where $\Gamma(x)$ is the Gamma function which for positive integers is equal to $\Gamma(n) = (n - 1)!$

And here are a few more that aren't required in the original post:

1. $(1 + 1 + 1)! = 6$
10. $\lfloor\sqrt{\sqrt{10 \times 10}} \rfloor + \sqrt{\lfloor\sqrt{10}\rfloor} = 6$
11. $\lfloor\sqrt{\sqrt{11 \times 11}} \rfloor + \sqrt{\lfloor\sqrt{11}\rfloor} = 6$
13. $\lfloor\sqrt{\sqrt{13 \times 13}} \rfloor + \sqrt{\lfloor\sqrt{13}\rfloor} = 6$
14. $\lfloor\sqrt{\sqrt{14 \times 14}} \rfloor + \sqrt{\lfloor\sqrt{14}\rfloor} = 6$
16. $\sqrt{16} + \sqrt{\sqrt{\sqrt{16 * 16}}} = 6$

• Finish it, don't google it. Sep 7, 2018 at 19:38
• Why is googling not allowed? If it is, please direct me to the part of the rules page where it says this. You asked for it! Sep 7, 2018 at 19:40
• hhhhh, if google is allowed, then what value to it ? The good thing is that google can't answer the last two rows. Sep 7, 2018 at 19:48
• @benasfan yes, I see that google isn't the best tool for finding these sorts of things - most only do 0 to 9. But, sometimes searching pays off. I found 12 online and then that gave me the insight to do 15. Sep 7, 2018 at 20:12
• @benasfan I see, but the floor function isn't that complicated... it results in 6: wolframalpha link Sep 7, 2018 at 22:44

After looking at @Hugh's answer, particularly the comments below it, I wanted to try and solve a few parts of the puzzle without using the internet. It is up to you whether or not you believe me.

The following is all I have found thus far.

2. $$2^2+2=6$$

3. $$3!+3-3=6\tag*{\big(n!=1\times 2\times \cdots \times n\big)}$$

4. $$4+4-\sqrt{4} = 6$$

5. $$\pi(5+5+5)=6\tag*{\big(\pi(x) = PCF\big)}$$ See here.

6. $$6^{6\,\div\, 6} = 6$$

7. $$\big\lceil \sqrt7+\sqrt{7+\sqrt7}\big\rceil=6$$ I could substitute $$7$$ with $$8$$ or $$9$$ and change the ceilng function $$\lceil\ldots\rceil$$ to the floor $$\lfloor\ldots\rfloor$$ instead and then have solutions regarding $$8$$ and $$9$$... but that would ruin some fun, I suppose.

8. $$\bigg\lfloor\frac{8+\sqrt{8}}{\sqrt{\sqrt{8}}}\bigg\rfloor=6$$ Also works for $$9$$ and if I change the ceiling function to the floor, it works for $$6$$ and $$7$$.

$$\sqrt{9}!\times 9\div 9 = 6$$ I would have instead added $$9$$ and then subtracted $$9$$ as opposed to multiplying and dividing, but I already did that with regards to $$3$$ and wanted to make this that little extra different.

10. $$\big\lfloor\sqrt{10}+\sqrt{10}\big\rfloor_{10}=6\tag*{\big(n_{10}=n\big)}$$

• Good for you for not googling, my first instinct in a question like this is to google, especially because this type of question is so standard. I assume that someone, somewhere on this planet has already solved a problem like this, they just need to be found. Sep 7, 2018 at 22:22
• I do believe you. No need to prove it's the truth or not. Sep 7, 2018 at 22:25
• @Hugh hahah, I know, right? That's what I think about when I see that the Twin Prime Conjecture still remains unproven... unless someone out there in the world has the proof but just hasn't submitted it or something, hahah :P Sep 7, 2018 at 22:56
• @user477343 this whole "Riemann hypothesis" thing? somebody must know the answer! lol Sep 7, 2018 at 23:01
• @user477343 Now do the math for 12 and 15 . U almost done. Sep 7, 2018 at 23:06