Mixing
Make 4 4 4 4 = 30,31
Clash Royale CLAN TAG#URR8PPP
up vote
15
down vote
favorite
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.
mathematics formation-of-numbers
edited Aug 29 at 14:03
Bass
22.2k355143
asked Aug 29 at 12:09
Oray
14.2k435140
 |Â
show 2 more comments
up vote
15
down vote
favorite
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.
mathematics formation-of-numbers
edited Aug 29 at 14:03
Bass
22.2k355143
asked Aug 29 at 12:09
Oray
14.2k435140
Is there any rule that we cannot change the rhs or touch it?
– R.D
Aug 29 at 12:12
@R.D nope, it is limited with LHS
– Oray
Aug 29 at 12:13
@Oray Can I reorder the numbers on the LHS?
– rhsquared
Aug 29 at 12:19
@rhsquared they are all the same though
– R.D
Aug 29 at 12:20
@R.D Doh! I can see that.
– rhsquared
Aug 29 at 12:23
 |Â
show 2 more comments
up vote
15
down vote
favorite
up vote
15
down vote
favorite
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.
mathematics formation-of-numbers
edited Aug 29 at 14:03
Bass
22.2k355143
asked Aug 29 at 12:09
Oray
14.2k435140
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.
mathematics formation-of-numbers
edited Aug 29 at 14:03
Bass
22.2k355143
asked Aug 29 at 12:09
Oray
14.2k435140
edited Aug 29 at 14:03
Bass
22.2k355143
edited Aug 29 at 14:03
Bass
22.2k355143
edited Aug 29 at 14:03
Bass
22.2k355143
22.2k355143
asked Aug 29 at 12:09
Oray
14.2k435140
asked Aug 29 at 12:09
Oray
14.2k435140
asked Aug 29 at 12:09
Oray
14.2k435140
14.2k435140
Is there any rule that we cannot change the rhs or touch it?
– R.D
Aug 29 at 12:12
@R.D nope, it is limited with LHS
– Oray
Aug 29 at 12:13
@Oray Can I reorder the numbers on the LHS?
– rhsquared
Aug 29 at 12:19
@rhsquared they are all the same though
– R.D
Aug 29 at 12:20
@R.D Doh! I can see that.
– rhsquared
Aug 29 at 12:23
 |Â
show 2 more comments
Is there any rule that we cannot change the rhs or touch it?
– R.D
Aug 29 at 12:12
@R.D nope, it is limited with LHS
– Oray
Aug 29 at 12:13
@Oray Can I reorder the numbers on the LHS?
– rhsquared
Aug 29 at 12:19
@rhsquared they are all the same though
– R.D
Aug 29 at 12:20
@R.D Doh! I can see that.
– rhsquared
Aug 29 at 12:23
Is there any rule that we cannot change the rhs or touch it?
– R.D
Aug 29 at 12:12
Is there any rule that we cannot change the rhs or touch it?
– R.D
Aug 29 at 12:12
@R.D nope, it is limited with LHS
– Oray
Aug 29 at 12:13
@R.D nope, it is limited with LHS
– Oray
Aug 29 at 12:13
@Oray Can I reorder the numbers on the LHS?
– rhsquared
Aug 29 at 12:19
@Oray Can I reorder the numbers on the LHS?
– rhsquared
Aug 29 at 12:19
@rhsquared they are all the same though
– R.D
Aug 29 at 12:20
@rhsquared they are all the same though
– R.D
Aug 29 at 12:20
@R.D Doh! I can see that.
– rhsquared
Aug 29 at 12:23
@R.D Doh! I can see that.
– rhsquared
Aug 29 at 12:23
 |Â
show 2 more comments
7 Answers
7
active
oldest
votes
up vote
21
down vote
accepted
For the first one
$(4 + (4/4))!/4 = 30$
Second one
$4! + (4!+4)/4 = 31$
answered Aug 29 at 12:14
hexomino
28.9k292144
2
it took too long! :)
– Oray
Aug 29 at 12:33
1
Oof. Beat me to the second one by just one minute
– R.D
Aug 29 at 12:33
add a comment |Â
up vote
8
down vote
Four Fours
FIRST:
$sqrt4 +sqrt4 +sqrt4 + 4!=30$
$(4times 4times sqrt4) -sqrt4 = 30$
$((4times 4!) + 4!)div 4 = 30$
$4! + sqrt4 + (4div sqrt4) = 30$
Really similar to 1:
$4! - sqrt4+4+4 = 30$
Really similar to 2:
$(4^sqrt4times sqrt4)-sqrt4= 30$
SECOND:
$((4+sqrt4)!+4!)div 4! = 31$
Solutions that bend the rules, slightly.
$(4+4+4)div .4 =30$
$4!+sqrt4+(sqrt4div .4) = 31$
Weird resemblance between these two other solutions!
$big(sqrtsqrtsqrt4^,4! - 4big)div sqrt4=30$
$big(sqrtsqrtsqrt4^,4! - sqrt4big)div sqrt4=31$
$$$$
Three Fours
FIRST:
$(4!div 4)+4!=30$
answered Aug 29 at 13:06
user477343
3,0601742
3
I think roots and log is not allowed here :P
– Ian Fako
Aug 29 at 13:10
@IanFako I forgot to read the most important part...
– user477343
Aug 29 at 13:11
1
Wow those nested radicals are awesome
– sedrick
Aug 29 at 13:37
1
@sedrick technically that's an eighth root $sqrt [8]4$ but I didn't write the eight.
– user477343
Aug 29 at 13:41
2
Isn't $4! - sqrt4 + (4div .4) = 32$? $(24) - (2) + (10)$
– JAD
Aug 30 at 7:55
 |Â
show 1 more comment
up vote
7
down vote
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(frac4!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-2uproot24 * 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 = biggllfloor sqrtsqrtsqrtsqrt(4!)! biggrrfloor$
$31 = biggllceil sqrtsqrtsqrtsqrt(4!)! biggrrceil$
answered Aug 29 at 13:21
sedrick
1,656514
1
@user477343 Man we're taking these puzzles waaaay too seriously
– sedrick
Aug 29 at 14:01
2
Oh my... the last answer is a beast!!
– user477343
Aug 29 at 14:06
1
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.
– sedrick
Aug 29 at 14:18
1
with JUST ONE 4 killed me. +1
– Aric
Aug 30 at 11:57
1
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.
– rosuav
Aug 30 at 12:25
 |Â
show 4 more comments
up vote
3
down vote
The second one (with double factorial)
$4!! * 4 - (4/4)$
answered Aug 29 at 12:22
Ian Fako
350112
!! is different operator than !
– Oray
Aug 29 at 12:23
He didn't mention double factorial though
– R.D
Aug 29 at 12:24
2
"by adding any symbols", it's not prohibited according to the question
– Ian Fako
Aug 29 at 12:25
Up to OP to decide if it's right or wrong XD
– R.D
Aug 29 at 12:26
1
The question does impose limit on symbols, but clearly states "any operations".
– Imre
Aug 30 at 7:36
 |Â
show 2 more comments
up vote
3
down vote
For the first one (Double factorial used)
$(4! + 4 + frac4!!4) = 30$
For the second one (Double factorial again):
$(4! + 4!! - frac4 4) = 31$
edited Aug 30 at 2:18
kraby15
2,2712730
answered Aug 29 at 13:10
A. MartÃn
311
add a comment |Â
up vote
1
down vote
For 30:
$(!4 - 4) times left( frac4!4 right)$
$ =(9 - 4) times left( frac244 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)^ii!$
As such
$!4 = 4! sum_i=0^4 frac(-1)^ii! = 24 times left(frac(-1)^00! + frac(-1)^11! + frac(-1)^22! + frac(-1)^33! + frac(-1)^44!right)$
$= 24 times left(frac11 + frac-11 + frac12 + frac-16 + frac124right)$
$= left(24 - 24 + 12 - 4 + 1right)$
$= 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
answered Aug 31 at 22:08
Jonathan Allan
17.1k14595
add a comment |Â
up vote
0
down vote
How About
4444 != 31
and
4444 != 30
Reason
!= means not equal to in many (programming) languages
answered Sep 5 at 9:22
Khushraj Rathod
31413
add a comment |Â
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
21
down vote
accepted
For the first one
$(4 + (4/4))!/4 = 30$
Second one
$4! + (4!+4)/4 = 31$
answered Aug 29 at 12:14
hexomino
28.9k292144
2
it took too long! :)
– Oray
Aug 29 at 12:33
1
Oof. Beat me to the second one by just one minute
– R.D
Aug 29 at 12:33
add a comment |Â
up vote
21
down vote
accepted
For the first one
$(4 + (4/4))!/4 = 30$
Second one
$4! + (4!+4)/4 = 31$
answered Aug 29 at 12:14
hexomino
28.9k292144
2
it took too long! :)
– Oray
Aug 29 at 12:33
1
Oof. Beat me to the second one by just one minute
– R.D
Aug 29 at 12:33
add a comment |Â
up vote
21
down vote
accepted
up vote
21
down vote
accepted
For the first one
$(4 + (4/4))!/4 = 30$
Second one
$4! + (4!+4)/4 = 31$
answered Aug 29 at 12:14
hexomino
28.9k292144
For the first one
$(4 + (4/4))!/4 = 30$
Second one
$4! + (4!+4)/4 = 31$
answered Aug 29 at 12:14
hexomino
28.9k292144
edited Aug 29 at 12:31
answered Aug 29 at 12:14
hexomino
28.9k292144
answered Aug 29 at 12:14
hexomino
28.9k292144
answered Aug 29 at 12:14
hexomino
28.9k292144
28.9k292144
2
it took too long! :)
– Oray
Aug 29 at 12:33
1
Oof. Beat me to the second one by just one minute
– R.D
Aug 29 at 12:33
add a comment |Â
2
it took too long! :)
– Oray
Aug 29 at 12:33
1
Oof. Beat me to the second one by just one minute
– R.D
Aug 29 at 12:33
2
2
it took too long! :)
– Oray
Aug 29 at 12:33
it took too long! :)
– Oray
Aug 29 at 12:33
1
1
Oof. Beat me to the second one by just one minute
– R.D
Aug 29 at 12:33
Oof. Beat me to the second one by just one minute
– R.D
Aug 29 at 12:33
add a comment |Â
up vote
8
down vote
Four Fours
FIRST:
$sqrt4 +sqrt4 +sqrt4 + 4!=30$
$(4times 4times sqrt4) -sqrt4 = 30$
$((4times 4!) + 4!)div 4 = 30$
$4! + sqrt4 + (4div sqrt4) = 30$
Really similar to 1:
$4! - sqrt4+4+4 = 30$
Really similar to 2:
$(4^sqrt4times sqrt4)-sqrt4= 30$
SECOND:
$((4+sqrt4)!+4!)div 4! = 31$
Solutions that bend the rules, slightly.
$(4+4+4)div .4 =30$
$4!+sqrt4+(sqrt4div .4) = 31$
Weird resemblance between these two other solutions!
$big(sqrtsqrtsqrt4^,4! - 4big)div sqrt4=30$
$big(sqrtsqrtsqrt4^,4! - sqrt4big)div sqrt4=31$
$$$$
Three Fours
FIRST:
$(4!div 4)+4!=30$
answered Aug 29 at 13:06
user477343
3,0601742
3
I think roots and log is not allowed here :P
– Ian Fako
Aug 29 at 13:10
@IanFako I forgot to read the most important part...
– user477343
Aug 29 at 13:11
1
Wow those nested radicals are awesome
– sedrick
Aug 29 at 13:37
1
@sedrick technically that's an eighth root $sqrt [8]4$ but I didn't write the eight.
– user477343
Aug 29 at 13:41
2
Isn't $4! - sqrt4 + (4div .4) = 32$? $(24) - (2) + (10)$
– JAD
Aug 30 at 7:55
 |Â
show 1 more comment
up vote
8
down vote
Four Fours
FIRST:
$sqrt4 +sqrt4 +sqrt4 + 4!=30$
$(4times 4times sqrt4) -sqrt4 = 30$
$((4times 4!) + 4!)div 4 = 30$
$4! + sqrt4 + (4div sqrt4) = 30$
Really similar to 1:
$4! - sqrt4+4+4 = 30$
Really similar to 2:
$(4^sqrt4times sqrt4)-sqrt4= 30$
SECOND:
$((4+sqrt4)!+4!)div 4! = 31$
Solutions that bend the rules, slightly.
$(4+4+4)div .4 =30$
$4!+sqrt4+(sqrt4div .4) = 31$
Weird resemblance between these two other solutions!
$big(sqrtsqrtsqrt4^,4! - 4big)div sqrt4=30$
$big(sqrtsqrtsqrt4^,4! - sqrt4big)div sqrt4=31$
$$$$
Three Fours
FIRST:
$(4!div 4)+4!=30$
answered Aug 29 at 13:06
user477343
3,0601742
3
I think roots and log is not allowed here :P
– Ian Fako
Aug 29 at 13:10
@IanFako I forgot to read the most important part...
– user477343
Aug 29 at 13:11
1
Wow those nested radicals are awesome
– sedrick
Aug 29 at 13:37
1
@sedrick technically that's an eighth root $sqrt [8]4$ but I didn't write the eight.
– user477343
Aug 29 at 13:41
2
Isn't $4! - sqrt4 + (4div .4) = 32$? $(24) - (2) + (10)$
– JAD
Aug 30 at 7:55
 |Â
show 1 more comment
up vote
8
down vote
up vote
8
down vote
Four Fours
FIRST:
$sqrt4 +sqrt4 +sqrt4 + 4!=30$
$(4times 4times sqrt4) -sqrt4 = 30$
$((4times 4!) + 4!)div 4 = 30$
$4! + sqrt4 + (4div sqrt4) = 30$
Really similar to 1:
$4! - sqrt4+4+4 = 30$
Really similar to 2:
$(4^sqrt4times sqrt4)-sqrt4= 30$
SECOND:
$((4+sqrt4)!+4!)div 4! = 31$
Solutions that bend the rules, slightly.
$(4+4+4)div .4 =30$
$4!+sqrt4+(sqrt4div .4) = 31$
Weird resemblance between these two other solutions!
$big(sqrtsqrtsqrt4^,4! - 4big)div sqrt4=30$
$big(sqrtsqrtsqrt4^,4! - sqrt4big)div sqrt4=31$
$$$$
Three Fours
FIRST:
$(4!div 4)+4!=30$
answered Aug 29 at 13:06
user477343
3,0601742
Four Fours
FIRST:
$sqrt4 +sqrt4 +sqrt4 + 4!=30$
$(4times 4times sqrt4) -sqrt4 = 30$
$((4times 4!) + 4!)div 4 = 30$
$4! + sqrt4 + (4div sqrt4) = 30$
Really similar to 1:
$4! - sqrt4+4+4 = 30$
Really similar to 2:
$(4^sqrt4times sqrt4)-sqrt4= 30$
SECOND:
$((4+sqrt4)!+4!)div 4! = 31$
Solutions that bend the rules, slightly.
$(4+4+4)div .4 =30$
$4!+sqrt4+(sqrt4div .4) = 31$
Weird resemblance between these two other solutions!
$big(sqrtsqrtsqrt4^,4! - 4big)div sqrt4=30$
$big(sqrtsqrtsqrt4^,4! - sqrt4big)div sqrt4=31$
$$$$
Three Fours
FIRST:
$(4!div 4)+4!=30$
answered Aug 29 at 13:06
user477343
3,0601742
edited Aug 31 at 0:18
answered Aug 29 at 13:06
user477343
3,0601742
answered Aug 29 at 13:06
user477343
3,0601742
answered Aug 29 at 13:06
user477343
3,0601742
3,0601742
3
I think roots and log is not allowed here :P
– Ian Fako
Aug 29 at 13:10
@IanFako I forgot to read the most important part...
– user477343
Aug 29 at 13:11
1
Wow those nested radicals are awesome
– sedrick
Aug 29 at 13:37
1
@sedrick technically that's an eighth root $sqrt [8]4$ but I didn't write the eight.
– user477343
Aug 29 at 13:41
2
Isn't $4! - sqrt4 + (4div .4) = 32$? $(24) - (2) + (10)$
– JAD
Aug 30 at 7:55
 |Â
show 1 more comment
3
I think roots and log is not allowed here :P
– Ian Fako
Aug 29 at 13:10
@IanFako I forgot to read the most important part...
– user477343
Aug 29 at 13:11
1
Wow those nested radicals are awesome
– sedrick
Aug 29 at 13:37
1
@sedrick technically that's an eighth root $sqrt [8]4$ but I didn't write the eight.
– user477343
Aug 29 at 13:41
2
Isn't $4! - sqrt4 + (4div .4) = 32$? $(24) - (2) + (10)$
– JAD
Aug 30 at 7:55
3
3
I think roots and log is not allowed here :P
– Ian Fako
Aug 29 at 13:10
I think roots and log is not allowed here :P
– Ian Fako
Aug 29 at 13:10
@IanFako I forgot to read the most important part...
– user477343
Aug 29 at 13:11
@IanFako I forgot to read the most important part...
– user477343
Aug 29 at 13:11
1
1
Wow those nested radicals are awesome
– sedrick
Aug 29 at 13:37
Wow those nested radicals are awesome
– sedrick
Aug 29 at 13:37
1
1
@sedrick technically that's an eighth root $sqrt [8]4$ but I didn't write the eight.
– user477343
Aug 29 at 13:41
@sedrick technically that's an eighth root $sqrt [8]4$ but I didn't write the eight.
– user477343
Aug 29 at 13:41
2
2
Isn't $4! - sqrt4 + (4div .4) = 32$? $(24) - (2) + (10)$
– JAD
Aug 30 at 7:55
Isn't $4! - sqrt4 + (4div .4) = 32$? $(24) - (2) + (10)$
– JAD
Aug 30 at 7:55
 |Â
show 1 more comment
up vote
7
down vote
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:
$frac(frac4!4)!4! = 30$
Making $31$ with just three $4$'s (violates rules):
$16$th root of $24!$ is $30.69$ so
$biggl lceil sqrt[leftroot-2uproot24 * 4](4!)! biggr rceil = 31$
Since we're already violating lots of rules in the first place, we can make both $30$ and $31$ with JUST ONE $4$.
$30 = biggllfloor sqrtsqrtsqrtsqrt(4!)! biggrrfloor$
$31 = biggllceil sqrtsqrtsqrtsqrt(4!)! biggrrceil$
answered Aug 29 at 13:21
sedrick
1,656514
1
@user477343 Man we're taking these puzzles waaaay too seriously
– sedrick
Aug 29 at 14:01
2
Oh my... the last answer is a beast!!
– user477343
Aug 29 at 14:06
1
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.
– sedrick
Aug 29 at 14:18
1
with JUST ONE 4 killed me. +1
– Aric
Aug 30 at 11:57
1
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.
– rosuav
Aug 30 at 12:25
 |Â
show 4 more comments
up vote
7
down vote
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:
$frac(frac4!4)!4! = 30$
Making $31$ with just three $4$'s (violates rules):
$16$th root of $24!$ is $30.69$ so
$biggl lceil sqrt[leftroot-2uproot24 * 4](4!)! biggr rceil = 31$
Since we're already violating lots of rules in the first place, we can make both $30$ and $31$ with JUST ONE $4$.
$30 = biggllfloor sqrtsqrtsqrtsqrt(4!)! biggrrfloor$
$31 = biggllceil sqrtsqrtsqrtsqrt(4!)! biggrrceil$
answered Aug 29 at 13:21
sedrick
1,656514
1
@user477343 Man we're taking these puzzles waaaay too seriously
– sedrick
Aug 29 at 14:01
2
Oh my... the last answer is a beast!!
– user477343
Aug 29 at 14:06
1
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.
– sedrick
Aug 29 at 14:18
1
with JUST ONE 4 killed me. +1
– Aric
Aug 30 at 11:57
1
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.
– rosuav
Aug 30 at 12:25
 |Â
show 4 more comments
up vote
7
down vote
up vote
7
down vote
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:
$frac(frac4!4)!4! = 30$
Making $31$ with just three $4$'s (violates rules):
$16$th root of $24!$ is $30.69$ so
$biggl lceil sqrt[leftroot-2uproot24 * 4](4!)! biggr rceil = 31$
Since we're already violating lots of rules in the first place, we can make both $30$ and $31$ with JUST ONE $4$.
$30 = biggllfloor sqrtsqrtsqrtsqrt(4!)! biggrrfloor$
$31 = biggllceil sqrtsqrtsqrtsqrt(4!)! biggrrceil$
answered Aug 29 at 13:21
sedrick
1,656514
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:
$frac(frac4!4)!4! = 30$
Making $31$ with just three $4$'s (violates rules):
$16$th root of $24!$ is $30.69$ so
$biggl lceil sqrt[leftroot-2uproot24 * 4](4!)! biggr rceil = 31$
Since we're already violating lots of rules in the first place, we can make both $30$ and $31$ with JUST ONE $4$.
$30 = biggllfloor sqrtsqrtsqrtsqrt(4!)! biggrrfloor$
$31 = biggllceil sqrtsqrtsqrtsqrt(4!)! biggrrceil$
answered Aug 29 at 13:21
sedrick
1,656514
edited Aug 29 at 13:59
answered Aug 29 at 13:21
sedrick
1,656514
answered Aug 29 at 13:21
sedrick
1,656514
answered Aug 29 at 13:21
sedrick
1,656514
1,656514
1
@user477343 Man we're taking these puzzles waaaay too seriously
– sedrick
Aug 29 at 14:01
2
Oh my... the last answer is a beast!!
– user477343
Aug 29 at 14:06
1
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.
– sedrick
Aug 29 at 14:18
1
with JUST ONE 4 killed me. +1
– Aric
Aug 30 at 11:57
1
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.
– rosuav
Aug 30 at 12:25
 |Â
show 4 more comments
1
@user477343 Man we're taking these puzzles waaaay too seriously
– sedrick
Aug 29 at 14:01
2
Oh my... the last answer is a beast!!
– user477343
Aug 29 at 14:06
1
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.
– sedrick
Aug 29 at 14:18
1
with JUST ONE 4 killed me. +1
– Aric
Aug 30 at 11:57
1
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.
– rosuav
Aug 30 at 12:25
1
1
@user477343 Man we're taking these puzzles waaaay too seriously
– sedrick
Aug 29 at 14:01
@user477343 Man we're taking these puzzles waaaay too seriously
– sedrick
Aug 29 at 14:01
2
2
Oh my... the last answer is a beast!!
– user477343
Aug 29 at 14:06
Oh my... the last answer is a beast!!
– user477343
Aug 29 at 14:06
1
1
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.
– sedrick
Aug 29 at 14:18
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.
– sedrick
Aug 29 at 14:18
1
1
with JUST ONE 4 killed me. +1
– Aric
Aug 30 at 11:57
with JUST ONE 4 killed me. +1
– Aric
Aug 30 at 11:57
1
1
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.
– rosuav
Aug 30 at 12:25
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.
– rosuav
Aug 30 at 12:25
 |Â
show 4 more comments
up vote
3
down vote
The second one (with double factorial)
$4!! * 4 - (4/4)$
answered Aug 29 at 12:22
Ian Fako
350112
!! is different operator than !
– Oray
Aug 29 at 12:23
He didn't mention double factorial though
– R.D
Aug 29 at 12:24
2
"by adding any symbols", it's not prohibited according to the question
– Ian Fako
Aug 29 at 12:25
Up to OP to decide if it's right or wrong XD
– R.D
Aug 29 at 12:26
1
The question does impose limit on symbols, but clearly states "any operations".
– Imre
Aug 30 at 7:36
 |Â
show 2 more comments
up vote
3
down vote
The second one (with double factorial)
$4!! * 4 - (4/4)$
answered Aug 29 at 12:22
Ian Fako
350112
!! is different operator than !
– Oray
Aug 29 at 12:23
He didn't mention double factorial though
– R.D
Aug 29 at 12:24
2
"by adding any symbols", it's not prohibited according to the question
– Ian Fako
Aug 29 at 12:25
Up to OP to decide if it's right or wrong XD
– R.D
Aug 29 at 12:26
1
The question does impose limit on symbols, but clearly states "any operations".
– Imre
Aug 30 at 7:36
 |Â
show 2 more comments
up vote
3
down vote
up vote
3
down vote
The second one (with double factorial)
$4!! * 4 - (4/4)$
answered Aug 29 at 12:22
Ian Fako
350112
The second one (with double factorial)
$4!! * 4 - (4/4)$
answered Aug 29 at 12:22
Ian Fako
350112
edited Aug 29 at 12:24
answered Aug 29 at 12:22
Ian Fako
350112
answered Aug 29 at 12:22
Ian Fako
350112
answered Aug 29 at 12:22
Ian Fako
350112
350112
!! is different operator than !
– Oray
Aug 29 at 12:23
He didn't mention double factorial though
– R.D
Aug 29 at 12:24
2
"by adding any symbols", it's not prohibited according to the question
– Ian Fako
Aug 29 at 12:25
Up to OP to decide if it's right or wrong XD
– R.D
Aug 29 at 12:26
1
The question does impose limit on symbols, but clearly states "any operations".
– Imre
Aug 30 at 7:36
 |Â
show 2 more comments
!! is different operator than !
– Oray
Aug 29 at 12:23
He didn't mention double factorial though
– R.D
Aug 29 at 12:24
2
"by adding any symbols", it's not prohibited according to the question
– Ian Fako
Aug 29 at 12:25
Up to OP to decide if it's right or wrong XD
– R.D
Aug 29 at 12:26
1
The question does impose limit on symbols, but clearly states "any operations".
– Imre
Aug 30 at 7:36
!! is different operator than !
– Oray
Aug 29 at 12:23
!! is different operator than !
– Oray
Aug 29 at 12:23
He didn't mention double factorial though
– R.D
Aug 29 at 12:24
He didn't mention double factorial though
– R.D
Aug 29 at 12:24
2
2
"by adding any symbols", it's not prohibited according to the question
– Ian Fako
Aug 29 at 12:25
"by adding any symbols", it's not prohibited according to the question
– Ian Fako
Aug 29 at 12:25
Up to OP to decide if it's right or wrong XD
– R.D
Aug 29 at 12:26
Up to OP to decide if it's right or wrong XD
– R.D
Aug 29 at 12:26
1
1
The question does impose limit on symbols, but clearly states "any operations".
– Imre
Aug 30 at 7:36
The question does impose limit on symbols, but clearly states "any operations".
– Imre
Aug 30 at 7:36
 |Â
show 2 more comments
up vote
3
down vote
For the first one (Double factorial used)
$(4! + 4 + frac4!!4) = 30$
For the second one (Double factorial again):
$(4! + 4!! - frac4 4) = 31$
edited Aug 30 at 2:18
kraby15
2,2712730
answered Aug 29 at 13:10
A. MartÃn
311
add a comment |Â
up vote
3
down vote
For the first one (Double factorial used)
$(4! + 4 + frac4!!4) = 30$
For the second one (Double factorial again):
$(4! + 4!! - frac4 4) = 31$
edited Aug 30 at 2:18
kraby15
2,2712730
answered Aug 29 at 13:10
A. MartÃn
311
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For the first one (Double factorial used)
$(4! + 4 + frac4!!4) = 30$
For the second one (Double factorial again):
$(4! + 4!! - frac4 4) = 31$
edited Aug 30 at 2:18
kraby15
2,2712730
answered Aug 29 at 13:10
A. MartÃn
311
For the first one (Double factorial used)
$(4! + 4 + frac4!!4) = 30$
For the second one (Double factorial again):
$(4! + 4!! - frac4 4) = 31$
edited Aug 30 at 2:18
kraby15
2,2712730
answered Aug 29 at 13:10
A. MartÃn
311
edited Aug 30 at 2:18
kraby15
2,2712730
edited Aug 30 at 2:18
kraby15
2,2712730
edited Aug 30 at 2:18
kraby15
2,2712730
2,2712730
answered Aug 29 at 13:10
A. MartÃn
311
answered Aug 29 at 13:10
A. MartÃn
311
answered Aug 29 at 13:10
A. MartÃn
311
311
add a comment |Â
add a comment |Â
up vote
1
down vote
For 30:
$(!4 - 4) times left( frac4!4 right)$
$ =(9 - 4) times left( frac244 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
$n! sum_i=0^n frac(-1)^ii!$
As such
$!4 = 4! sum_i=0^4 frac(-1)^ii! = 24 times left(frac(-1)^00! + frac(-1)^11! + frac(-1)^22! + frac(-1)^33! + frac(-1)^44!right)$
$= 24 times left(frac11 + frac-11 + frac12 + frac-16 + frac124right)$
$= left(24 - 24 + 12 - 4 + 1right)$
$= 9$
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
answered Aug 31 at 22:08
Jonathan Allan
17.1k14595
add a comment |Â
up vote
1
down vote
For 30:
$(!4 - 4) times left( frac4!4 right)$
$ =(9 - 4) times left( frac244 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
$n! sum_i=0^n frac(-1)^ii!$
As such
$!4 = 4! sum_i=0^4 frac(-1)^ii! = 24 times left(frac(-1)^00! + frac(-1)^11! + frac(-1)^22! + frac(-1)^33! + frac(-1)^44!right)$
$= 24 times left(frac11 + frac-11 + frac12 + frac-16 + frac124right)$
$= left(24 - 24 + 12 - 4 + 1right)$
$= 9$
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
answered Aug 31 at 22:08
Jonathan Allan
17.1k14595
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For 30:
$(!4 - 4) times left( frac4!4 right)$
$ =(9 - 4) times left( frac244 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
$n! sum_i=0^n frac(-1)^ii!$
As such
$!4 = 4! sum_i=0^4 frac(-1)^ii! = 24 times left(frac(-1)^00! + frac(-1)^11! + frac(-1)^22! + frac(-1)^33! + frac(-1)^44!right)$
$= 24 times left(frac11 + frac-11 + frac12 + frac-16 + frac124right)$
$= left(24 - 24 + 12 - 4 + 1right)$
$= 9$
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
answered Aug 31 at 22:08
Jonathan Allan
17.1k14595
For 30:
$(!4 - 4) times left( frac4!4 right)$
$ =(9 - 4) times left( frac244 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
$n! sum_i=0^n frac(-1)^ii!$
As such
$!4 = 4! sum_i=0^4 frac(-1)^ii! = 24 times left(frac(-1)^00! + frac(-1)^11! + frac(-1)^22! + frac(-1)^33! + frac(-1)^44!right)$
$= 24 times left(frac11 + frac-11 + frac12 + frac-16 + frac124right)$
$= left(24 - 24 + 12 - 4 + 1right)$
$= 9$
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
answered Aug 31 at 22:08
Jonathan Allan
17.1k14595
edited Aug 31 at 22:18
answered Aug 31 at 22:08
Jonathan Allan
17.1k14595
answered Aug 31 at 22:08
Jonathan Allan
17.1k14595
answered Aug 31 at 22:08
Jonathan Allan
17.1k14595
17.1k14595
add a comment |Â
add a comment |Â
up vote
0
down vote
How About
4444 != 31
and
4444 != 30
Reason
!= means not equal to in many (programming) languages
answered Sep 5 at 9:22
Khushraj Rathod
31413
add a comment |Â
up vote
0
down vote
How About
4444 != 31
and
4444 != 30
Reason
!= means not equal to in many (programming) languages
answered Sep 5 at 9:22
Khushraj Rathod
31413
add a comment |Â
up vote
0
down vote
up vote
0
down vote
How About
4444 != 31
and
4444 != 30
Reason
!= means not equal to in many (programming) languages
answered Sep 5 at 9:22
Khushraj Rathod
31413
How About
4444 != 31
and
4444 != 30
Reason
!= means not equal to in many (programming) languages
answered Sep 5 at 9:22
Khushraj Rathod
31413
answered Sep 5 at 9:22
Khushraj Rathod
31413
answered Sep 5 at 9:22
Khushraj Rathod
31413
answered Sep 5 at 9:22
Khushraj Rathod
31413
31413
add a comment |Â
add a comment |Â
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Is there any rule that we cannot change the rhs or touch it?
– R.D
Aug 29 at 12:12
@R.D nope, it is limited with LHS
– Oray
Aug 29 at 12:13
@Oray Can I reorder the numbers on the LHS?
– rhsquared
Aug 29 at 12:19
@rhsquared they are all the same though
– R.D
Aug 29 at 12:20
@R.D Doh! I can see that.
– rhsquared
Aug 29 at 12:23