8
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So the 6s challenge consist of taking a number from 0 to 10 and, using only "common" operations and the number three times, obtaining the number 6.

For example: $(0!+0!+0!)! = 6$ and $(4 - (4/4))! = 6$.

Now, for all the numbers from 0 to 7 you can find a solution that doesn't involve using any roots (check this video for possible solutions: https://www.youtube.com/watch?v=h2vkrxvh76c), but every solution I've came across, for 8,9 and 10 uses them.

For example: $8 - \sqrt(\sqrt(8+8)) = 6$ and $(\sqrt(9))! + 9-9 = 6$

My question is if you know any solution for 8,9 and 10 that uses only $+,-,\cdot,\div,(,)$ and $!$.

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  • $\begingroup$ I'm pretty certain there's no "rootless" solution with these rules. Even if we allowed for writing the digits together, using the decimal point, and exponentiation, it seems very difficult. $\endgroup$ – Bass Jun 26 at 22:20
6
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A possible solution for 8:

If I can use !! with interpretation as double factorial, we have $((8!!)/8)/8 = ((8\cdot6\cdot4\cdot2)/8)/8=6$

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4
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Solution using $10$:

$\big((10!!)_!\big)_!-\frac{10}{\Big(\big((10!!)_!\big)_!\Big)_!} = (3840_!)_!-\frac{10}{\big((3840_!)_!\big)_!} = 38_!-\frac{10}{(38_!)_!} = 11-\frac{10}{11_!} = 11-\frac{10}{2} = 11-5=6$ where $(x)_!$ is the factorial number system.

Another solution using $10$ which I don't regard as an answer though (a silly one) is:

$10+10+10 = 6$ (binary $2+2+2=6$)

Solution using two $9$s:

$$\Bigg(\frac{(9!!)_!}{9}\Bigg)! = \Bigg(\frac{945_!}{9}\Bigg)!=\Bigg(\frac{27}{9}\Bigg)!=6$$

If we are allowed to use another mathematical operator $\%$ (modulus, a close relative of division) then the solution using three $9$s is:

$$\Bigg(\frac{(9!!)_!}{9}\Bigg)!\%9 = \Bigg(\frac{945_!}{9}\Bigg)!\%9=\Bigg(\frac{27}{9}\Bigg)!\%9=6\%9=6$$

Solution using just one mathematical operator $\bar{}$ ($1$'s complement) if allowed using $9$

$\bar{9} = \overline{1001} = 0110 = 6$

Leading to three $9$s solution using two operations:

$\bar{9}+9-9 = 6$

Also, an interesting thing to note is that

$6 = 9$ (XS3 code) or using three $9$s as exprected $6=9+9-9 $

Another solution using $10$

$\Big(-\big(-(10!!)_{-10}\big)_{-10}\Big)_{-!} = \Big(-\big(-3840_{-10}\big)_{-10}\Big)_{-!}= \Big(-2240_{-10}\Big)_{-!} = 1840_{-!} = 1\times (-1)^3\times 3! + 8\times (-1)^2 \times 2! + 4\times (-1)^1\times 1! = 6$ where $(x)_{-10}$ is the negadecimal system and $(x)_{-!}$ is the negative factorial system.

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  • 1
    $\begingroup$ That's only two 9s. $\endgroup$ – Herb Wolfe Jun 27 at 0:32

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