# Use math symbols to produce 6 [duplicate]

This question already has an answer here:

How can we get (6) from these numbers?

$2$ $2$ $2 = 6$

$3$ $3$ $3 = 6$

$4$ $4$ $4 = 6$

$5$ $5$ $5 = 6$

$6$ $6$ $6 = 6$

$7$ $7$ $7 = 6$

$8$ $8$ $8 = 6$

$9$ $9$ $9 = 6$

$12$ $12$ $12 = 6$

$15$ $15$ $15 = 6$

## marked as duplicate by JonMark Perry, Quintec, El-Guest, DqwertyC, Jaap ScherphuisSep 7 '18 at 23:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 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)? – Hugh Sep 7 '18 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? – Hugh Sep 7 '18 at 22:20
• if you want you could carry the concept on indefinitely, but the principle is the same as in the duplicate. @Hugh – JonMark Perry Sep 7 '18 at 22:30
• @benasfan is my answer better now? – Hugh Sep 8 '18 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. – ben asfan Sep 8 '18 at 18:45

## 2 Answers

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. – ben asfan Sep 7 '18 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! – Hugh Sep 7 '18 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. – ben asfan Sep 7 '18 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. – Hugh Sep 7 '18 at 20:12
• @benasfan I see, but the floor function isn't that complicated... it results in 6: wolframalpha link – Hugh Sep 7 '18 at 22:44

# Partial Answer:

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. – Hugh Sep 7 '18 at 22:22
• I do believe you. No need to prove it's the truth or not. – ben asfan Sep 7 '18 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 – Feeds Sep 7 '18 at 22:56
• @user477343 this whole "Riemann hypothesis" thing? somebody must know the answer! lol – Hugh Sep 7 '18 at 23:01
• @user477343 Now do the math for 12 and 15 . U almost done. – ben asfan Sep 7 '18 at 23:06