21
$\begingroup$

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 $\sqrt{9}! - \frac{9}{9} = 5$

  • 1
    In $\dfrac{9}{9}$ the 9 is used only twice.2012-06-02
  • 7
    They should have multiplied $\dfrac 99$ with $0.\overline 9$.2012-06-02
  • 2
    Or write $\dfrac{\sqrt{9}\sqrt{9}}{9}$2012-06-02
  • 3
    @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 5.2012-06-02
  • 0
    @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..2012-06-02
  • 3
    @Phira It's not wrong at 5 if you see it like: (sqrt(9))! - 9/9 = 52012-06-02
  • 0
    @Lipis The main difficulty seems to be how to replace $.\overline{9}=1$.2012-06-02
  • 4
    @Lipis but it does say $\sqrt{9!}-\frac{9}{9}$.2012-06-02
  • 1
    Related: http://math.stackexchange.com/questions/16066/designing-an-irrational-numbers-wall-clock2012-06-02
  • 0
    @Thomas agree.. the guy who was copying from the paper did a mistake before making it a clock ;)2012-06-02
  • 0
    Do you allow $\log$?2012-06-02
  • 0
    @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`2012-06-02
  • 1
    @Lipis If you allow logs and square roots, then you can write any integer with two 2s.2012-06-02
  • 0
    @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 :)2012-06-02
  • 0
    @Lipis You can also use the logs for other $n$, you just need to go to $k=4$ to divide by the log of 2.2012-06-02
  • 2
    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.2012-06-02
  • 1
    @AméricoTavares Yes, thanks for that.. I updated my question.. sorry :(2012-06-02
  • 1
    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.2012-06-02
  • 0
    @LieRyan if you can come up with these answers.. would be nice :)2012-06-02
  • 0
    I think that this might work well as community wiki.2012-06-02
  • 1
    tzador's contributions are incredible!2012-06-03
  • 0
    You kinda CAN just rounding, but with the floor function. I get the point of the question though2015-11-22

14 Answers 14

14

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

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

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

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

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

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

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

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

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

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

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

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

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

  • 2
    guys, help me validate this2012-06-02
  • 19
    While this is impressive I'm not sure how'd it would go on a clock.2012-06-02
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).

  • 0
    I think the problem with square roots is that they're implicitly using twos.2012-06-02
  • 0
    Without rounding as long as it's valid.2012-06-02
  • 0
    @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.2012-06-02
  • 0
    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.2012-06-02
  • 0
    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.2012-06-02
  • 0
    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 :)2012-06-02
  • 1
    point taken. :)2012-06-02
  • 0
    @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.2012-06-02
  • 0
    Or ninth roots in the clock in the picture,2012-06-03
11

solution for n = 1, k = 12:

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

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

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

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

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

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

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

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

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

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

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

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

  • 0
    This is one way to adjust [Evan Carroll's answer](http://math.stackexchange.com/a/152945/1424) to better fit the question's specifications.2012-06-02
  • 1
    This also has the nice property of having lines of identical length2012-06-02
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 $3$.

  • 0
    Any suggestions on how to improve $11$?2012-06-02
  • 0
    Well.. the problem is that you're using the number 2 not exactly `k` times to produce each number, as described in the problem.2012-06-02
  • 0
    $((2*2)!-2)/2=11$2012-06-02
  • 0
    @Phira that only has $4$ twos2012-06-02
  • 1
    $(2*2)!/2-2/2$ if you prefer2012-06-02
  • 1
    And $10=22/2-2/2$2012-06-02
  • 0
    @Lipis Looks like we're done here.2012-06-02
  • 0
    Can someone edit the question and use LaTeX for the expressions and = 1, 2, 3, etc ;)2012-06-02
  • 2
    Your 3. has only four 2s. $2 + 22/22$ would do.2012-06-02
  • 0
    @PeterPhipps thanks for catching that.2012-06-02
  • 0
    A less complicated 11: $\frac{22}{2}\times\frac{2}{2}$.2012-06-02
  • 0
    I think you can use another answer for the 5s.. Just to keep one answer per pair..2012-06-02
  • 2
    Are you saying that $1=2-2/2+2/2=2-1+1=2$? Does that mean that I don't have to write my thesis because you found a contradiction in mathematics? Awesome.2012-06-02
  • 0
    @AsafKaragila Lol. You might just be able to start writing your thesis. I fixed it btw.2012-06-02
  • 2
    I don't like those 22. $3 = \frac{(2+2)!}{(2*2*2)}$ $10 = \frac{(2^{2+2})}{2} +2$2012-06-02
  • 0
    Well they fall within the parameters of the question.2012-06-02
  • 0
    Well, the question ask "numbers" and not digits. But the 9 clock uses 99. So, probably the question is a bit wrong. Anyway, i just said i didn't like them! ;)2012-06-02
  • 1
    4 doesn't seem to work, I get 6. There are plenty of other solutions though: I like $ {2^2}^2 / 2^2$2012-06-02
6

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

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

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

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

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

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

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

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

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

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

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

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

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

5

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

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

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

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

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

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

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

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

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

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

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

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

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

3

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

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

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

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

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

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

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

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

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

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

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

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

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

  • 3
    One wonders how this would go on a clock.2012-06-02
  • 2
    @Eugene: Not only one, even two, three and four ...2012-06-02
3

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

$1 = 3^{3-3}$

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

$3 = 3+3-3$

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

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

$6 = 3*3-3$

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

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

$9 = 3+3+3$

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

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

$12 = 3*3+3$

  • 0
    What is $\pi(3)$? It's 2 in your case.. but any reference?2012-06-04
  • 2
    It's the prime-counting function. $\pi(x)$ is the number of prime numbers equal or less than x.2012-06-04
  • 0
    Awesome...! Thanks2012-06-04
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 $9$.

  • 3
    $\left(5+\frac{5!}{\left(5+\left(5 \times 5\right)\right)}\right)=9$2012-06-02
  • 3
    $9=\left(5-\frac{\left(5-\left(5 \times 5\right)\right)}{5}\right)$ what about this one?2012-06-02
  • 0
    I don't think 55 counts as two instances of five; it's a separate number, after all.2012-06-02
  • 1
    You should look at the original clock.2012-06-02
2

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

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

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

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

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

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

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

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

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

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

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

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

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

  • 0
    actuall for k in [1..12] and n = 12 there is a solutions2012-06-02
2

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

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

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

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

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

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

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

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

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

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

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

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

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

  • 0
    I will update my question to include negative numbers as well.. why not?2012-06-02
2

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

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

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

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

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

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

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

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

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

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

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

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

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

0

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.

  • 0
    What's your `n` and `k`?2012-06-02
  • 0
    "I don't think five ones is possible."--That is interesting. What restrictions do you have in mind for what operations/functions are allowed?2012-06-02
  • 0
    $n = 1, k = 6$; 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 $k = 5$.)2012-06-02
  • 0
    @JonasMeyer Any number as long as the expression uses only that number as is. For `n=14` the number `a=141414` also allowed..2012-06-02
  • 1
    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.2012-06-02
-3

For $1$ here is a solution:

$$ 1 = 1 $$

$$ 1+1 = 2 $$

$$ 1+1+1 = 3 $$

$$ 1+1+1+1 = 4 $$

$$ 1+1+1+1+1 = 5 $$

$$ 1+1+1+1+1+1 = 6 $$

$$ 1+1+1+1+1+1+1 = 7 $$

$$ 1+1+1+1+1+1+1+1 = 8 $$

$$ 1+1+1+1+1+1+1+1+1 = 9 $$

$$ 1+1+1+1+1+1+1+1+1+1 = 10 $$

$$ 1+1+1+1+1+1+1+1+1+1+1 = 11 $$

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

  • 7
    1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 9, 10 = 10, 11 = 11, 12 = 122012-06-02
  • 1
    We are to fix n and k...2012-06-02
  • 0
    I don't understand the downvotes. It's trivial to fix, e.g. $8=1\cdot 1\cdot1\cdot1(1+1+1+1+1+1+1+1)$, $n=1$ & $k=12$.2012-06-02
  • 1
    @JonasMeyer: Except that was my answer2012-06-02
  • 1
    Oh, apparently someone else posted that as a new answer. instead. (Edit) @Eric: Yes, I see that now. It wasn't your answer until 9 minutes ago, and I hadn't seen it.2012-06-02