15
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Follow up question to Make 5 5 5 5 = 19


Can you find a way to make:

$4\ 4 \ 4 \ 4 = 30$

and

$4\ 4 \ 4 \ 4 = 31$

by adding any operations or symbols on the left side of the equations? You can use only these symbols:

$+,\ -,\ *,\ !,\ /,\ \hat\, ,\ ()$.

It is limited to this list, and concatenation is also allowed.

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  • $\begingroup$ Is there any rule that we cannot change the rhs or touch it? $\endgroup$ – R.D Aug 29 '18 at 12:12
  • $\begingroup$ @R.D nope, it is limited with LHS $\endgroup$ – Oray Aug 29 '18 at 12:13
  • $\begingroup$ @Oray Can I reorder the numbers on the LHS? $\endgroup$ – rhsquared Aug 29 '18 at 12:19
  • $\begingroup$ @rhsquared they are all the same though $\endgroup$ – R.D Aug 29 '18 at 12:20
  • $\begingroup$ @R.D Doh! I can see that. $\endgroup$ – rhsquared Aug 29 '18 at 12:23
21
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For the first one

$(4 + (4/4))!/4 = 30$

Second one

$4! + (4!+4)/4 = 31$

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  • 2
    $\begingroup$ it took too long! :) $\endgroup$ – Oray Aug 29 '18 at 12:33
  • 1
    $\begingroup$ Oof. Beat me to the second one by just one minute $\endgroup$ – R.D Aug 29 '18 at 12:33
9
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Four Fours

FIRST:

  1. $\sqrt4 +\sqrt4 +\sqrt4 + 4!=30$

  1. $(4\times 4\times \sqrt{4}) -\sqrt{4} = 30$

  1. $((4\times 4!) + 4!)\div 4 = 30$

  1. $4! + \sqrt{4} + (4\div \sqrt{4}) = 30$

Really similar to 1:

$4! - \sqrt{4}+4+4 = 30$

Really similar to 2:

$(4^{\sqrt{4}}\times \sqrt{4})-\sqrt{4}= 30$

SECOND:

  1. $((4+\sqrt{4})!+4!)\div 4! = 31$


Solutions that bend the rules, slightly.

  1. $(4+4+4)\div .4 =30$

  1. $4!+\sqrt{4}+(\sqrt{4}\div .4) = 31$


Weird resemblance between these two other solutions!

$\big(\sqrt{\sqrt{\sqrt{4}}}^{\,4!} - 4\big)\div \sqrt{4}=30$

$\big(\sqrt{\sqrt{\sqrt{4}}}^{\,4!} - \sqrt4\big)\div \sqrt{4}=31$

$$$$

Three Fours

FIRST:

$(4!\div 4)+4!=30$

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  • 3
    $\begingroup$ I think roots and log is not allowed here :P $\endgroup$ – Ian Fako Aug 29 '18 at 13:10
  • $\begingroup$ @IanFako I forgot to read the most important part... $\endgroup$ – Feeds Aug 29 '18 at 13:11
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    $\begingroup$ Wow those nested radicals are awesome $\endgroup$ – sedrick Aug 29 '18 at 13:37
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    $\begingroup$ @sedrick technically that's an eighth root $\sqrt [8]{4}$ but I didn't write the eight. $\endgroup$ – Feeds Aug 29 '18 at 13:41
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    $\begingroup$ Isn't $4! - \sqrt{4} + (4\div .4) = 32$? $(24) - (2) + (10)$ $\endgroup$ – JAD Aug 30 '18 at 7:55
7
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Another possible answer:

$(4 - (4/4))! + 4! = 30$

A weird but fun stretch answer:

If you concatenate $4/4$ and $4!$ that's $124$.
$124 / 4 = 31$

Making $30$ with just three $4$'s:

$\frac{(\frac{4!}{4})!}{4!} = 30$

Making $31$ with just three $4$'s (violates rules):

$16$th root of $24!$ is $30.69$ so
$\biggl \lceil \sqrt[\leftroot{-2}\uproot{2}{4 * 4}]{(4!)!} \biggr \rceil = 31$

Since we're already violating lots of rules in the first place, we can make both $30$ and $31$ with JUST ONE $4$.

$30 = \biggl\lfloor \sqrt{\sqrt{\sqrt{\sqrt{(4!)!}}}} \biggr\rfloor$
$31 = \biggl\lceil \sqrt{\sqrt{\sqrt{\sqrt{(4!)!}}}} \biggr\rceil$

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  • 1
    $\begingroup$ @user477343 Man we're taking these puzzles waaaay too seriously $\endgroup$ – sedrick Aug 29 '18 at 14:01
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    $\begingroup$ Oh my... the last answer is a beast!! $\endgroup$ – Feeds Aug 29 '18 at 14:06
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    $\begingroup$ So I got curious after finding that last answer and did some research. This is pretty interesting math.stackexchange.com/questions/48633/… It's apparently possible to start with a single 4 and end with any positive integer. $\endgroup$ – sedrick Aug 29 '18 at 14:18
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    $\begingroup$ with JUST ONE 4 killed me. +1 $\endgroup$ – Aric Aug 30 '18 at 11:57
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    $\begingroup$ Upvoted just for the ridiculous "one four" solution. I wonder if you can make an insane RISC computer capable of just these two arithmetic operations, and define all of mathematics in terms of them. $\endgroup$ – rosuav Aug 30 '18 at 12:25
3
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The second one (with double factorial)

$4!! * 4 - (4/4)$

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  • $\begingroup$ !! is different operator than ! $\endgroup$ – Oray Aug 29 '18 at 12:23
  • $\begingroup$ He didn't mention double factorial though $\endgroup$ – R.D Aug 29 '18 at 12:24
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    $\begingroup$ "by adding any symbols", it's not prohibited according to the question $\endgroup$ – Ian Fako Aug 29 '18 at 12:25
  • $\begingroup$ Up to OP to decide if it's right or wrong XD $\endgroup$ – R.D Aug 29 '18 at 12:26
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    $\begingroup$ The question does impose limit on symbols, but clearly states "any operations". $\endgroup$ – Imre Aug 30 '18 at 7:36
3
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For the first one (Double factorial used)

$(4! + 4 + \frac{4!!}{4}) = 30$

For the second one (Double factorial again):

$(4! + 4!! - \frac{4 }{ 4}) = 31$

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1
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For 30:

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

$ =(9 - 4) \times \left( \frac{24}{4} \right) = 5 \times 6 = 30$

For 31:

$44 - 4 - !4$

$= 44 - 4 - 9 = 31$

Note that:

$!n$ is the subfactorial of $n$.
For a non-negative-integer $n$ this is the number of derangements of $n$
(the number of ways to arrange $n$ items such that no item is at its naturally ordered position)
This is
$n! \sum_{i=0}^n \frac{(-1)^i}{i!}$

As such
$!4 = 4! \sum_{i=0}^4 \frac{(-1)^i}{i!} = 24 \times \left(\frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!}\right)$
$= 24 \times \left(\frac{1}{1} + \frac{-1}{1} + \frac{1}{2} + \frac{-1}{6} + \frac{1}{24}\right)$
$= \left(24 - 24 + 12 - 4 + 1\right)$
$= 9$

Or, using ABCD, the 9 derangements are:
1. BADC
2. BCDA
3. BDAC
4. CADB
5. CDAB
6. CDBA
7. DABC
8. DCAB
9. DCBA

But not any of the other 15 permutations:
ABCD . ACDB . BACD . CABD . DACB
ABDC . ADBC . BCAD . CBAD . DBAC
ACBD . ADCB . BDCA . CBDA . DBCA

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0
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How About

4444 != 31

and

4444 != 30

Reason

!= means not equal to in many (programming) languages

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