Make 4 4 4 4 = 30,31

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.

• Is there any rule that we cannot change the rhs or touch it?
– R.D
Commented Aug 29, 2018 at 12:12
• @R.D nope, it is limited with LHS
– Oray
Commented Aug 29, 2018 at 12:13
• @Oray Can I reorder the numbers on the LHS? Commented Aug 29, 2018 at 12:19
• @rhsquared they are all the same though
– R.D
Commented Aug 29, 2018 at 12:20
• @R.D Doh! I can see that. Commented Aug 29, 2018 at 12:23

For the first one

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

Second one

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

• it took too long! :)
– Oray
Commented Aug 29, 2018 at 12:33
• Oof. Beat me to the second one by just one minute
– R.D
Commented Aug 29, 2018 at 12:33

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:

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

• I think roots and log is not allowed here :P Commented Aug 29, 2018 at 13:10
• @IanFako I forgot to read the most important part... Commented Aug 29, 2018 at 13:11
• Wow those nested radicals are awesome Commented Aug 29, 2018 at 13:37
• @sedrick technically that's an eighth root $\sqrt [8]{4}$ but I didn't write the eight. Commented Aug 29, 2018 at 13:41
• Isn't $4! - \sqrt{4} + (4\div .4) = 32$? $(24) - (2) + (10)$
Commented Aug 30, 2018 at 7:55

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

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 $4$.

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

• @user477343 Man we're taking these puzzles waaaay too seriously Commented Aug 29, 2018 at 14:01
• Oh my... the last answer is a beast!! Commented Aug 29, 2018 at 14:06
• 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. Commented Aug 29, 2018 at 14:18
• with JUST ONE 4 killed me. +1
– Aric
Commented Aug 30, 2018 at 11:57
• 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. Commented Aug 30, 2018 at 12:25

The second one (with double factorial)

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

• !! is different operator than !
– Oray
Commented Aug 29, 2018 at 12:23
• He didn't mention double factorial though
– R.D
Commented Aug 29, 2018 at 12:24
• "by adding any symbols", it's not prohibited according to the question Commented Aug 29, 2018 at 12:25
• Up to OP to decide if it's right or wrong XD
– R.D
Commented Aug 29, 2018 at 12:26
• The question does impose limit on symbols, but clearly states "any operations".
– Imre
Commented Aug 30, 2018 at 7:36

For the first one (Double factorial used)

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

For the second one (Double factorial again):

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

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:
2. BCDA
3. BDAC
5. CDAB
6. CDBA
7. DABC
8. DCAB
9. DCBA

But not any of the other 15 permutations:
ABCD . ACDB . BACD . CABD . DACB