18
$\begingroup$

Can you use the digits 0,0,0 and 1 each only once in a mathematical expression for the number 5 using only common mathematical symbols with at least 1 mathematical symbol between each number?

$\endgroup$
35
$\begingroup$

Let's go with

$(1!+0!+0!)!-0!=5$

$\endgroup$
  • 4
    $\begingroup$ Good answer, but I think you can use just (1) instead of 1 factorial (1!) $\endgroup$ – Sherlock Holmes May 28 '17 at 10:35
  • $\begingroup$ Parallelism or whatever the word it is. $\endgroup$ – bleh May 28 '17 at 19:56
  • $\begingroup$ @SherlockHolmes Wouldn't look as pretty, though $\endgroup$ – Rob May 29 '17 at 5:30
  • 2
    $\begingroup$ @bleh Vectorization $\endgroup$ – Aiman Al-Eryani May 29 '17 at 10:27
  • 1
    $\begingroup$ @AimanAl-Eryani No, it is parrallelism $\endgroup$ – Wen1now May 29 '17 at 11:02
41
$\begingroup$

First off, latest edit - just for fun, how to get 5 from just 0 and 1:

$$\left\lceil\sqrt{\left\lceil\sqrt{\frac{0!}{.1}}\right\rceil!}\right\rceil$$

Before rule change posted:

$$\frac{10}{0! + 0!}$$

With the changed rules:

$$\frac{\frac{0!}{.1}}{0! + 0!}$$

And while we're at it, here's $0$ to $28$:

$$0 \cdot 0 \cdot 0 \cdot 1 = 0$$
$$0 \cdot 0 \cdot 0 + 1 = 1$$
$$0 \cdot 0 + 0! + 1 = 2$$
$$0 + 0! + 0! + 1 = 3$$
$$0! + 0! + 0! + 1 = 4$$
$$\frac{\frac{0!}{.1}}{0! + 0!} = 5$$
$$1\cdot(0! + 0! + 0!)! = 6$$
$$1 + (0! + 0! + 0!)! = 7$$
$$\frac{0!}{.1} - 0! - 0! = 8$$
$$\frac{0!}{.1} - 0! \cdot 0! = 9$$
$$\frac{0! \cdot 0! \cdot 0!}{.1} = 10$$
$$\frac{0!}{.1} + 0! \cdot 0! = 11$$
$$\frac{0!}{.1} + 0! + 0! = 12$$
$$\left\lfloor\frac{\sqrt{0! + 0!}}{.1}\right\rfloor - 0! =13$$
$$\left\lfloor\frac{\sqrt{0! + 0!}}{.1}\right\rfloor \cdot 0! =14$$
$$\left\lfloor\frac{\sqrt{0! + 0!}}{.1}\right\rfloor + 0! =15$$
$$\left\lceil\sqrt{\frac{0!}{.1}}\right\rceil ^ {0! + 0!} = 16$$
$$\left\lfloor\frac{\sqrt{0! + 0! +0!}}{.1}\right\rfloor = 17$$
$$\left\lceil\frac{\sqrt{0! + 0! +0!}}{.1}\right\rceil = 18$$
$$\frac{0! + 0!}{.1} - 0! = 19$$
$$\frac{0! + 0!}{.1} \cdot 0! = 20$$
$$\frac{0! + 0!}{.1} + 0! = 21$$
$$\left\lceil\sqrt{\frac{0!}{.1}}\right\rceil! - 0! - 0! = 22$$
$$\left\lceil\sqrt{\frac{0!}{.1}}\right\rceil! - 0! \cdot 0! = 23$$
$$(1 + 0! + 0! + 0!)! = 24$$
$$\left\lceil\sqrt{\frac{0!}{.1}}\right\rceil! + 0! \cdot 0! = 25$$
$$\left\lceil\sqrt{(((0! + 0! + 0!)!)!}\right\rceil - 0! = 26$$
$$\left\lceil\sqrt{(((0! + 0! + 0!)!)!}\right\rceil \cdot 0! = 27$$
$$\left\lceil\sqrt{(((0! + 0! + 0!)!)!}\right\rceil + 0! = 28$$

And here's how to get 5 from just 0, 0 and 1:

$$\left\lceil\sqrt{\frac{0!}{.1}}\right\rceil + 0!$$

And how to get 5 from just 0, 0 and 0:

$$\left\lfloor\sqrt{\sqrt{((0! + 0! + 0!)!)!}}\right\rfloor$$

$\endgroup$
  • $\begingroup$ Sorry, I realised I should have changed my edit earlier. Your answer works for my original question, but can you do the new one? $\endgroup$ – Y. Zhang May 28 '17 at 9:59
  • $\begingroup$ This program just prints a hardcoded equation! Why using a program? just write it by hand. A nice program would output the equation given the desired output and the allowed terms and operators. $\endgroup$ – Sembei Norimaki May 30 '17 at 8:59
  • $\begingroup$ @SembeiNorimaki fixed it for you :) $\endgroup$ – Paul Evans May 30 '17 at 13:15
  • $\begingroup$ make sure you are taking the factorials one by one, otherwise you could imply the double factorial. :) $\endgroup$ – Quintec Jun 2 '17 at 23:51
  • $\begingroup$ @thecoder16 thanks - edited accordingly :) $\endgroup$ – Paul Evans Jun 2 '17 at 23:55
22
$\begingroup$

A simple variation with factorials:

$$(0!+0!+0!)!-1 = 5$$

$\endgroup$
  • 1
    $\begingroup$ While this is, arguably, aesthetically prettier,  it is only a trivial variation on the accepted answer. $\endgroup$ – Peregrine Rook May 29 '17 at 17:48
  • 30
    $\begingroup$ While it is, arguably, only a trivial variation on the accepted answer, it is aesthetically prettier. $\endgroup$ – Florian F May 29 '17 at 19:44
5
$\begingroup$

Just for fun

${\Big\lceil} \sqrt{(0! + 0! + 0! + 1!)!} \> \Big\rceil$

or

${\Huge\lfloor} \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{((0!+0!+0!+1!)!)!}}}}}{\Huge\rfloor}$

$\endgroup$
4
$\begingroup$

As a derivative

$ \frac {d}{dx}[((0!+0!+0!)!)-1]x $

If the question was changed to require NO mathematical symbols, then using Base 5, we could write

0010 = (0*5^3 + 0*5^2 +1*5^1 +0*5^0)

In fact we could write any positive integer n using base n as 0010!

$\endgroup$
  • 3
    $\begingroup$ Well representing n as 10 (base n) is kinda trivially cheating... works for any positive integer n... but you still have to construct the '5' from somewhere. $\endgroup$ – smci May 30 '17 at 7:25
  • $\begingroup$ @smci I don't see how my comment about base 5 is "cheating," since it doesn't even answer the question (which specified there must be mathematical symbols between each digit). I highlighted this with a capital "NO" in my post. If anything its off topic. However, the fact that it works for any n certainly does not make it cheating. Now, if you want to argue that converting 0010 to base to requires a 5 to multiply the 1, I agree. However as a representation of 5 in base 5 I see no problem. $\endgroup$ – Evan Rosica Jun 2 '17 at 23:32
  • $\begingroup$ @smci Unless you want to argue that denoting it as base 5 requires a 5, in which case I reply that all answers must denote base 10, and thus are incorrect since they use too many 1s and 0s..... $\endgroup$ – Evan Rosica Jun 2 '17 at 23:35
  • $\begingroup$ Evan, ok I see the second part is an offtopic sidenote; you might make it say so more clearly. I wasn't attacking you, sir. Just noting that modification makes all these digit-arithmetic-manipulation genre questions utterly moot... As to whether we can understand '10' to be implicit in our (human-biased) decimal representation, yes. There has to be some default. I suppose we can argue that base-2 would be most universal species-independent choice, also allows for most efficient encoding (if our muscles allow us to have arbitrary adjacent fingers up and down...) $\endgroup$ – smci Jun 4 '17 at 20:43

protected by Beastly Gerbil Jun 11 '17 at 17:28

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.