# 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
Aug 29 '18 at 12:12
• @R.D nope, it is limited with LHS
– Oray
Aug 29 '18 at 12:13
• @Oray Can I reorder the numbers on the LHS? Aug 29 '18 at 12:19
• @rhsquared they are all the same though
– R.D
Aug 29 '18 at 12:20
• @R.D Doh! I can see that. Aug 29 '18 at 12:23

For the first one

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

Second one

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

• it took too long! :)
– Oray
Aug 29 '18 at 12:33
• Oof. Beat me to the second one by just one minute
– R.D
Aug 29 '18 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 Aug 29 '18 at 13:10
• @IanFako I forgot to read the most important part... Aug 29 '18 at 13:11
• Wow those nested radicals are awesome Aug 29 '18 at 13:37
• @sedrick technically that's an eighth root $\sqrt [8]{4}$ but I didn't write the eight. Aug 29 '18 at 13:41
• Isn't $4! - \sqrt{4} + (4\div .4) = 32$? $(24) - (2) + (10)$
Aug 30 '18 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 Aug 29 '18 at 14:01
• Oh my... the last answer is a beast!! Aug 29 '18 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. Aug 29 '18 at 14:18
• with JUST ONE 4 killed me. +1
– Aric
Aug 30 '18 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. Aug 30 '18 at 12:25

The second one (with double factorial)

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

• !! is different operator than !
– Oray
Aug 29 '18 at 12:23
• He didn't mention double factorial though
– R.D
Aug 29 '18 at 12:24
• "by adding any symbols", it's not prohibited according to the question Aug 29 '18 at 12:25
• Up to OP to decide if it's right or wrong XD
– R.D
Aug 29 '18 at 12:26
• The question does impose limit on symbols, but clearly states "any operations".
– Imre
Aug 30 '18 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