How can we produce another geek clock with a different pair of numbers?

Question

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

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 2

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 $3$.

Answer 3

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).

Answer 4

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 $9$.

Answer 5

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 6

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 7

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 8

solution for n = 1, k = 12:

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

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

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

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

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

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

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

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

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

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

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

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

Answer 9

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 10

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.

Answer 11

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 12

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 13

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

$1 = 3^{3-3}$

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

$3 = 3+3-3$

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

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

$6 = 3*3-3$

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

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

$9 = 3+3+3$

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

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

$12 = 3*3+3$