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So I found this geek clock and I think that it's pretty cool.

Geeky clock

I'm just wondering if it is possible to achieve the same but with another number.

So here is the problem:

We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok.

If you're answering with an example then use one pair per answer.

I just want to see that clock with another pair of numbers :)

Notes for the current clock:

1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.

5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$

Lipis
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    In $\dfrac{9}{9}$ the 9 is used only twice. – Américo Tavares Jun 02 '12 at 13:40
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    They should have multiplied $\dfrac 99$ with $0.\overline 9$. – Phira Jun 02 '12 at 13:43
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    Or write $\dfrac{\sqrt{9}\sqrt{9}}{9}$ – Américo Tavares Jun 02 '12 at 13:43
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    @Lipis This is not very hard. If you just want to make some small examples, I would recommend doing it by hand. Anyway, I just noticed that the clock is wrong at 5. – Phira Jun 02 '12 at 13:46
  • @AméricoTavares true... :) Didn't notice it.. but the problem remains the same :) Because here you could use the .9 repeating (which is the same with: 1 + 1 - 1) like in the number 7.. – Lipis Jun 02 '12 at 13:46
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    @Phira It's not wrong at 5 if you see it like: (sqrt(9))! - 9/9 = 5 – Lipis Jun 02 '12 at 13:47
  • @Lipis The main difficulty seems to be how to replace $.\overline{9}=1$. – Américo Tavares Jun 02 '12 at 13:51
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    @Lipis but it does say $\sqrt{9!}-\frac{9}{9}$. –  Jun 02 '12 at 13:52
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    Related: http://math.stackexchange.com/questions/16066/designing-an-irrational-numbers-wall-clock – Asaf Karagila Jun 02 '12 at 13:52
  • @Thomas agree.. the guy who was copying from the paper did a mistake before making it a clock ;) – Lipis Jun 02 '12 at 13:54
  • Do you allow $\log$? – Phira Jun 02 '12 at 14:01
  • @Phira sure.. whatever you can imagine.. as long as the result is a natural number.. and we can see the number `n` written `k` times :) it can also be like: `(99 / 99 + 9)` for `n = 9` and `k = 5` for the number `10` – Lipis Jun 02 '12 at 14:05
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    @Lipis If you allow logs and square roots, then you can write any integer with two 2s. – Phira Jun 02 '12 at 14:34
  • @Phira agree.. that solves it for `n = 2` then :) I want to see it also with a different pair of numbers.. not only the extreme ones :) – Lipis Jun 02 '12 at 14:36
  • @Lipis You can also use the logs for other $n$, you just need to go to $k=4$ to divide by the log of 2. – Phira Jun 02 '12 at 14:49
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    I deleted my suggestion (example with $n=1,k=4$) because I used the ceiling and the floor functions and I got the following comment: "No rounding.. sorry :(". Please add this constraint to the question. – Américo Tavares Jun 02 '12 at 15:14
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    @AméricoTavares Yes, thanks for that.. I updated my question.. sorry :( – Lipis Jun 02 '12 at 15:17
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    I guess the next question is, to find the set of all 2-tuples where this clock has a solution. Also, find a minimal set of operations that would make any (n,k) pair of naturals to have a solution. – Lie Ryan Jun 02 '12 at 19:27
  • @LieRyan if you can come up with these answers.. would be nice :) – Lipis Jun 02 '12 at 19:31
  • I think that this might work well as community wiki. – Jonas Meyer Jun 02 '12 at 22:57
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    tzador's contributions are incredible! – Samuel Reid Jun 03 '12 at 23:17
  • You kinda CAN just rounding, but with the floor function. I get the point of the question though –  Nov 22 '15 at 17:29

13 Answers13

14

For $n=12$ and $k=12$ here is a solution:

$1=\frac{12}{12+12+12+12+12+12-(12+12+12+12+12)}$

$2=\left(12 \times \frac{12}{12-12+12-12+12+12+12+12+12+12}\right)$

$3=\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)$

$4=\left(12-\frac{12}{\left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$5=\left(12 \times \frac{12}{\left(12 \times \left(12 \times \frac{12}{\left(12-\left(12-\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)}\right)\right)}\right)$

$6=\left(12+\frac{12}{\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$7=\frac{12}{\left(12 \times \frac{12}{\left(12-\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$8=\left(12+\left(12 \times \frac{12}{\left(12+\left(12+\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)\right)$

$9=\frac{12}{\left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$10=\left(12 \times \frac{12}{\left(12-\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)}\right)\right)}\right)$

$11=\left(12+\frac{12}{\left(12-\left(12 \times \left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)}\right)\right)\right)}\right)$

$12=\left(12+\left(12+\left(12-\left(12 \times \left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)}\right)\right)\right)\right)\right)$

tzador
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11

Making numbers out of 4 fours is a common problem: $$1=\frac {44}{44}$$ $$2=\frac {4\cdot 4}{4+4}$$ $$3=\frac{4+4+4}{4}$$ $$4=\frac{4-4}{4}+4$$ $$5=\sqrt{4!+\frac{\sqrt 4+\sqrt 4} 4}$$ $$6=\sqrt{\frac{4!\cdot 4-4!}{\sqrt 4}}$$ $$7=\sqrt{4!\sqrt 4+\frac 4 4}$$ $$8=\sqrt{\frac{4^4}{\sqrt4+\sqrt 4}}$$ $$9=(4-\frac 4 4)^{\sqrt 4}$$ $$10=\frac{4!} 4 - (4-\sqrt 4)$$ $$11=\frac{4!}{\sqrt 4}-\frac 4 4$$ $$12=\sqrt{\frac{4!4!}{\sqrt 4+\sqrt 4}}$$

You should clarify what operations you want. If you allow for any kind of rounding function, factorials and logs you can almost certainly do it with one of any number (though the resulting expressions may not fit on a clock).

Robert Mastragostino
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  • I think the problem with square roots is that they're implicitly using twos. – Eugene Jun 02 '12 at 14:57
  • Without rounding as long as it's valid. – Lipis Jun 02 '12 at 15:00
  • @Eugene the $\sqrt 9$s in the clock are then implicitly $3$s, and the $4!$ is implicitly 24, no? I'm not really arguing; I can try an minimize my use of them but I assumed they were allowed given that they appeared in the original clock. – Robert Mastragostino Jun 02 '12 at 15:05
  • I think given the original parameters of the question it's valid. I was just thinking that square roots use twos implicitly. Wasn't trying to deride your fine answer. Sorry if it came off that way. – Eugene Jun 02 '12 at 15:06
  • not at all. I was originally trying to only do it with $+,-,\times,\divide$, but it wasn't very fruitful. I like the challenge though: I'll try to minimize how often I square root a 4 directly. – Robert Mastragostino Jun 02 '12 at 15:08
  • Since there are multiple solutions.. I think it would be nicer to keep the cooler one and drop the other one :) Square roots are more impressive :) – Lipis Jun 02 '12 at 18:31
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  • @RobertMastragostino I think the point Eugene was trying to make was that square roots use 2s in the same way as cube roots use 3s. – Lily Chung Jun 02 '12 at 20:36
  • Or ninth roots in the clock in the picture, – Eugene Jun 03 '12 at 02:28
11

solution for n = 1, k = 12:

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1 = 2 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1 = 3 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1 = 4 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1 = 5 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1 = 6 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1 = 7 $$

$$ 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1 = 8 $$

$$ 1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1+1 = 9 $$

$$ 1 \times 1 \times 1+1+1+1+1+1+1+1+1+1 = 10 $$

$$ 1 \times 1+1+1+1+1+1+1+1+1+1+1 = 11 $$

$$ 1+1+1+1+1+1+1+1+1+1+1+1 = 12 $$

Eric
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  • This is one way to adjust [Evan Carroll's answer](http://math.stackexchange.com/a/152945/1424) to better fit the question's specifications. – Jonas Meyer Jun 02 '12 at 23:10
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    This also has the nice property of having lines of identical length – Eric Jun 02 '12 at 23:12
10

Seems like $2$ would do it:

$$ 1: 2^2 - 2 - 2/2 $$

$$ 2:2^2 - 2^2 + 2 $$

$$ 3: 2 + 22/22 $$

$$ 4: 2^{2^2}/2^2 $$

$$ 5: 2^2 - 2/2 + 2 $$

$$ 6: 2^2 + 2 - 2 + 2 $$

$$ 7:2^2 + 2 + 2/2 $$

$$ 8:2^{2}(2) + 2 - 2 $$

$$ 9:2^2(2) + 2/2 $$

$$ 10:22/2 - 2/2 $$

$$ 11 : (2^2)!/2 - 2/2 $$

$$ 12: 2^{2^2} - 2^2 $$

That should do it. Thanks to Phira for $10$ and $11$ and Peter for $3$.

Eugene
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6

For $n=9$ and $k=9$ here is a solution:

$1=\left(9+\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+\left(9-99\right)\right)\right)}\right)}\right)$

$2=\frac{9}{\left(9+\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{99}\right)}\right)}\right)}$

$3=\left(9-\left(9+\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)\right)\right)$

$4=\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+\left(9+\left(9+9\right)\right)\right)\right)}\right)}\right)$

$5=\left(9+\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+99\right)\right)}\right)}\right)\right)$

$6=\left(9+\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)}\right)$

$7=\left(9+\left(9+\frac{9}{\left(9 \times \frac{9}{\left(9-\left(9+99\right)\right)}\right)}\right)\right)$

$8=\left(9+\left(9 \times \frac{9}{\left(9+\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)\right)$

$9=\left(9 \times \left(9 \times \frac{9}{\left(9-\left(9+\left(9+\left(9-99\right)\right)\right)\right)}\right)\right)$

$10=\left(9-\left(9 \times \frac{9}{\left(9+\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)\right)$

$11=\frac{9}{\left(9 \times \frac{9}{\left(9-\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)}$

$12=\left(9-\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)}\right)$

tzador
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5

For $n=4$ and $k=5$ here is a solution:

$\frac{4}{\left(4+\left(4 \times \left(4-4\right)\right)\right)}=1$

$\left(4-\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=2$

$\left(4+\frac{4}{\left(4-\left(4+4\right)\right)}\right)=3$

$\left(4+\left(4+\left(4-\left(4+4\right)\right)\right)\right)=4$

$\left(4-\frac{4}{\left(4-\left(4+4\right)\right)}\right)=5$

$\left(4+\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=6$

$\frac{4}{\left(4 \times \frac{4}{\left(4+4!\right)}\right)}=7$

$\left(4 \times \left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=8$

$\left(4-\left(\frac{4}{4}-\frac{4!}{4}\right)\right)=9$

$\left(4+\frac{4}{\left(4 \times \frac{4}{4!}\right)}\right)=10$

$\frac{4}{\left(4 \times \frac{4}{44}\right)}=11$

$\left(4-\left(4-\left(4+\left(4+4\right)\right)\right)\right)=12$

tzador
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3

For $n=19$ and $k=19$ here is a solution:

$1=\frac{19}{\left(19+\left(19 \times \left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}$

$2=\frac{19}{\left(19 \times \frac{19}{\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$3=\left(19-\left(19+\frac{19}{\left(19-\left(19 \times \frac{19}{\left(19+\left(19 \times \frac{19}{\left(19+\left(19+\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)\right)}\right)\right)}\right)\right)$

$4=\left(19-\frac{19}{\left(19 \times \frac{19}{\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$5=\left(19 \times \frac{19}{\left(19+\left(19+\left(19+\left(19+\left(19 \times \frac{19}{\left(19+\left(19+\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)\right)\right)}\right)$

$6=\left(19+\left(19-\left(19+\frac{19}{\left(19 \times \frac{19}{\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)}\right)\right)\right)$

$7=\left(19-\left(19 \times \frac{19}{\left(19-\left(19 \times \frac{19}{\left(19+\left(19 \times \left(19 \times \frac{19}{\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)}\right)\right)}\right)\right)$

$8=\left(19+\frac{19}{\left(19 \times \frac{19}{\left(19+\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$9=\left(19 \times \frac{19}{\left(19+\left(19-\left(19 \times \frac{19}{\left(19+\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)}\right)$

$10=\frac{19}{\left(19 \times \frac{19}{\left(19-\left(19+\left(19+\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$11=\left(19+\frac{19}{\left(19-\left(19 \times \frac{19}{\left(19+\left(19 \times \frac{19}{\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)\right)}\right)\right)}\right)$

$12=\frac{19}{\left(19 \times \frac{19}{\left(19-\left(19+\left(19-\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+\left(19+19\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

tzador
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3

For $n=3$ and $k = 3$.

$1 = 3^{3-3}$

$2 = 3-\frac{3}{3}$

$3 = 3+3-3$

$4 = 3+\frac{3}{3}$

$5 = 3!-\frac{3}{3}$

$6 = 3*3-3$

$7 = 3!+\frac{3}{3}$

$8 = \pi(3)*\pi(3)*\pi(3)$

$9 = 3+3+3$

$10 = 3!+\pi(3)+\pi(3)$

$11 = 3!+3+\pi(3)$

$12 = 3*3+3$

2

Now with $n = 5$ and $k = 5$.

With $n = 5$ and $k = 5$ (missing a $9$ for now but I'll come back to it later).

$\dfrac{55}{5}-5-5=1$

$\dfrac{5+5}{5}-5+5=2$

$\dfrac{5+5}{5}+\frac{5}{5}=3$

$\dfrac{5+5+5+5}{5}=4$

$5 - 5 + 5 - 5 + 5 = 5$

$5 + \dfrac{5}{5} - 5 + 5 = 6$

$5 + \dfrac{5}{5}+\dfrac{5}{5} = 7$

$5 + 5 - \dfrac{5+5}{5} = 8$

$5 + \dfrac{5(5) - 5}{5}=9$

$\dfrac{55}{5} - \dfrac{5}{5} = 10$

$\dfrac{55}{5} - 5 + 5 = 11$

$\dfrac{5+5}{5} + 5 + 5 = 12$

Thanks to tzador for $9$.

Eugene
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2

For $n=2$ and $k=12$ here is a solution:

$1=\left(2 \times \left(2 \times \left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)$

$2=\left(2+\left(2 \times \left(2+\left(2+\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$

$3=\left(2 \times \left(2+\left(2 \times \frac{2}{\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)$

$4=\frac{2}{\left(2 \times \left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)}$

$5=\frac{2}{\left(2 \times \frac{2}{\left(2-\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$6=\left(2-\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$7=\frac{2}{\left(2 \times \frac{2}{\left(2-\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$8=\left(2-\left(2+\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$

$9=\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$10=\left(2+\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$11=\left(2+\left(2+\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)}\right)}\right)\right)$

$12=\left(2-\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$

tzador
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2

For $n=-1$ and $k=8$ here is a solution:

$1=\left(-1-\left(-1 \times \left(-1+\left(-1-\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$

$2=\left(-1+\left(-1+\left(-1-\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$

$3=\left(-1-\left(-1 \times \left(-1-\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$

$4=\left(-1+\left(-1-\left(-1+\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$

$5=\left(-1+\left(-1 \times \left(-1+\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$

$6=\left(-1-\left(-1+\left(-1+\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$

$7=\left(-1 \times \left(-1+\left(-1+\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$

$8=\left(-1 \times \left(-1+\left(-1-\left(\left(-1+-1\right) \times \left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)$

$9=\left(-1 \times \left(-1-\left(\left(-1+-1\right) \times \left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)$

$10=\left(-1 \times \left(-1-\left(\left(-1+\left(-1+-1\right)\right) \times \left(-1+\left(-1+-1\right)\right)\right)\right)\right)$

$11=\left(-1-\left(\left(-1+-1\right) \times \left(\left(-1+-1\right) \times \left(-1+\left(-1+-1\right)\right)\right)\right)\right)$

$12=\left(-1 \times \left(\left(-1+-1\right) \times \left(\left(-1+-1\right) \times \left(-1+\left(-1+-1\right)\right)\right)\right)\right)$

tzador
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2

or $n=-12$ and $k=12$ here is a solution:

$1=\frac{-12}{\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12-\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)}$

$2=\left(-12 \times \frac{-12}{\left(-12+\left(-12-\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)$

$3=\left(-12 \times \frac{-12}{\left(-12+\left(-12+\left(-12-\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)$

$4=\left(-12-\left(-12 \times \frac{-12}{\left(-12+\left(-12 \times \frac{-12}{\left(-12-\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)}\right)\right)}\right)\right)$

$5=\left(-12-\left(-12+\frac{-12}{\left(-12 \times \frac{-12}{\left(-12-\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)}\right)}\right)\right)$

$6=\left(-12+\left(-12 \times \left(-12 \times \frac{-12}{\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)$

$7=\frac{-12}{\left(-12 \times \frac{-12}{\left(-12-\left(-12-\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$8=\left(-12-\left(-12+\left(-12+\left(-12 \times \frac{-12}{\left(-12+\left(-12-\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)}\right)\right)\right)\right)$

$9=\frac{-12}{\left(-12 \times \frac{-12}{\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$10=\left(-12 \times \frac{-12}{\left(-12-\left(-12+\left(-12+\left(-12 \times \frac{-12}{\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)}\right)\right)\right)\right)}\right)$

$11=\left(-12-\frac{-12}{\left(-12 \times \frac{-12}{\left(-12-\left(-12 \times \left(-12+\left(-12-\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$12=\left(-12-\left(-12 \times \left(-12 \times \frac{-12}{\left(-12-\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+\left(-12+-12\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)$

tzador
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Six ones. For ease of reading, I write $n$ for the sum of $n$ 1s, so for example I mean

$1 = ((1+1+1)-(1+1)) \times 1$.

Here are the expressions:

$1 = (3-2) \times 1$

$2 = 4-2$

$3 = (4-1) \times 1$

$4 = 5-1$

$5 = 5 \times 1$

$6 = 6$

$7 = (3 \times 2) + 1$

$8 = 4 \times 2$

$9 = 3 \times 3$

$10 = (3! - 1) \times 2$

$11 = (3! \times 2) - 1$

$12 = 3! \times 2 \times 1$

Any larger number of 1s is possible (just multiply these expressions by 1 as many times as necessary); I don't think five ones is possible.

Michael Lugo
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  • What's your `n` and `k`? – Lipis Jun 02 '12 at 23:03
  • "I don't think five ones is possible."--That is interesting. What restrictions do you have in mind for what operations/functions are allowed? – Jonas Meyer Jun 02 '12 at 23:07
  • $n = 1, k = 6$; I had in mind addition, subtraction, multiplication, division, exponentiation, square roots, and factorials. (To be honest I didn't try that hard to get down to $k = 5$.) – Michael Lugo Jun 02 '12 at 23:09
  • @JonasMeyer Any number as long as the expression uses only that number as is. For `n=14` the number `a=141414` also allowed.. – Lipis Jun 02 '12 at 23:10
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    If we allow $11$ (using two 1s) then $n = 1$, $k = 5$ is possible. I suppose allowing concatenation is implicit in the way the original clock is set up. – Michael Lugo Jun 02 '12 at 23:12