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Make 24 using exactly three 3s
How many consecutive positive integers can you make using exactly four instances of the digit '4'?A truly amazing way of making the number 2016Use six 6s in an expression that equals 1,000A Wonderfully Tricky way of Making any Positive IntegerExpressing numbers using 0, 1, 2, 3, and 4Making 1-50 from 2016Most consecutive positive integers using two 1sMaking 4 with 4 ones (with a twist!)Make numbers 93 using the digits 2, 0, 1, 8Use 2, 0, 1 and 8 to make 71
$begingroup$
Make 24 using exactly three 3s
Each number formed with a 3 and the 24 in the equation are all base 10.
You cannot introduce any additional digits or constants.
Plus(+), negation/subtraction (-), division(/ or show it with a fraction bar), multiplication(* or use grouping symbols as appropriate), exponentiation(^ or show it with an exponent). regular factorial(!), square root(sqrt(), or the radical symbol), grouping symbols such as parentheses and/or square brackets, and the regular decimal point(not for the repeating decimal) are allowed. No concatenation of any type is allowed.
No other functions/operators are allowed. This includes floor and ceiling functions not being allowed. (edit)
I am looking for a total of 10 solutions.
formation-of-numbers arithmetic
asked May 22 at 21:03
Olive StemfornOlive Stemforn
733517
$endgroup$
add a comment |
$begingroup$
Make 24 using exactly three 3s
Each number formed with a 3 and the 24 in the equation are all base 10.
You cannot introduce any additional digits or constants.
Plus(+), negation/subtraction (-), division(/ or show it with a fraction bar), multiplication(* or use grouping symbols as appropriate), exponentiation(^ or show it with an exponent). regular factorial(!), square root(sqrt(), or the radical symbol), grouping symbols such as parentheses and/or square brackets, and the regular decimal point(not for the repeating decimal) are allowed. No concatenation of any type is allowed.
No other functions/operators are allowed. This includes floor and ceiling functions not being allowed. (edit)
I am looking for a total of 10 solutions.
formation-of-numbers arithmetic
asked May 22 at 21:03
Olive StemfornOlive Stemforn
733517
$endgroup$
6
$begingroup$
Let's please not make "non-serious" or otherwise very clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers.
$endgroup$
– Rubio♦
May 23 at 14:34
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
May 28 at 4:30
add a comment |
$begingroup$
Make 24 using exactly three 3s
Each number formed with a 3 and the 24 in the equation are all base 10.
You cannot introduce any additional digits or constants.
Plus(+), negation/subtraction (-), division(/ or show it with a fraction bar), multiplication(* or use grouping symbols as appropriate), exponentiation(^ or show it with an exponent). regular factorial(!), square root(sqrt(), or the radical symbol), grouping symbols such as parentheses and/or square brackets, and the regular decimal point(not for the repeating decimal) are allowed. No concatenation of any type is allowed.
No other functions/operators are allowed. This includes floor and ceiling functions not being allowed. (edit)
I am looking for a total of 10 solutions.
formation-of-numbers arithmetic
asked May 22 at 21:03
Olive StemfornOlive Stemforn
733517
$endgroup$
Make 24 using exactly three 3s
Each number formed with a 3 and the 24 in the equation are all base 10.
You cannot introduce any additional digits or constants.
Plus(+), negation/subtraction (-), division(/ or show it with a fraction bar), multiplication(* or use grouping symbols as appropriate), exponentiation(^ or show it with an exponent). regular factorial(!), square root(sqrt(), or the radical symbol), grouping symbols such as parentheses and/or square brackets, and the regular decimal point(not for the repeating decimal) are allowed. No concatenation of any type is allowed.
No other functions/operators are allowed. This includes floor and ceiling functions not being allowed. (edit)
I am looking for a total of 10 solutions.
formation-of-numbers arithmetic
formation-of-numbers arithmetic
asked May 22 at 21:03
Olive StemfornOlive Stemforn
733517
asked May 22 at 21:03
Olive StemfornOlive Stemforn
733517
edited May 23 at 19:25
Olive Stemforn
asked May 22 at 21:03
Olive StemfornOlive Stemforn
733517
asked May 22 at 21:03
Olive StemfornOlive Stemforn
733517
asked May 22 at 21:03
Olive StemfornOlive Stemforn
733517
733517
6
$begingroup$
Let's please not make "non-serious" or otherwise very clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers.
$endgroup$
– Rubio♦
May 23 at 14:34
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
May 28 at 4:30
add a comment |
6
$begingroup$
Let's please not make "non-serious" or otherwise very clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers.
$endgroup$
– Rubio♦
May 23 at 14:34
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
May 28 at 4:30
6
6
$begingroup$
Let's please not make "non-serious" or otherwise very clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers.
$endgroup$
– Rubio♦
May 23 at 14:34
$begingroup$
Let's please not make "non-serious" or otherwise very clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers.
$endgroup$
– Rubio♦
May 23 at 14:34
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
May 28 at 4:30
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
May 28 at 4:30
add a comment |
8 Answers
8
active
oldest
votes
$begingroup$
Ten solutions.
(i) $sqrt3^3! -3 = 24$
(ii) $left(frac3! + 3!3 right)! = 24$
(iii) $ left(frac3.3 - 3! right)! = 24$
(iv) $left(sqrtfrac3.3 + 3! right)! = 24$
(v) $left(3! - frac3!3 right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac33)! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt3)^3! -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8481935
answered May 22 at 21:29
hexominohexomino
52.2k4154246
$endgroup$
2
$begingroup$
@JonathanAllan That's what Olive's comment above was referring to.
$endgroup$
– hexomino
May 22 at 23:31
1
$begingroup$
@JonathanAllan the similar variant $(3 + fracsqrt3sqrt3)! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
4
$begingroup$
I'd say that i and ix are basically the same since the order of applying the square root and the power are irrelevant in the same way that 3x2 and 2x3 are the same (though that is just my subjective opinion).
$endgroup$
– Chris
May 23 at 14:45
1
$begingroup$
@Chris Yes I think you are right here, not sure if we really can consider them to be different considering what has been disallowed previously.
$endgroup$
– hexomino
May 23 at 15:32
2
$begingroup$
I see, and since it's more of a numerical "fluke" (in that it would only work for 3) that they cancel that stops them being classed as identical I guess.
$endgroup$
– Fifth_H0r5eman
May 23 at 15:33
|
show 19 more comments
$begingroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
14.5k33084
answered May 22 at 21:05
Olive StemfornOlive Stemforn
733517
$endgroup$
16
$begingroup$
This solution came to mind after reading just the title of the question. I assumed it was the only one.
$endgroup$
– JollyJoker
May 23 at 7:32
add a comment |
$begingroup$
Here is my first:
$(3 + frac33)! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
3,7231118
$endgroup$
1
$begingroup$
@ Weather Vane - Let us say that by adding the square root symbols, you have not changed the values of those. They are redundant. It's similar to writing 1! instead of 1 or 2! instead of 2. Although mathematically correct, I would not see them as "primitive solutions."
$endgroup$
– Olive Stemforn
May 22 at 21:44
1
$begingroup$
Ah OK, I was about to add another similar variant - removed.
$endgroup$
– Weather Vane
May 22 at 21:45
add a comment |
$begingroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,0301646
$endgroup$
2
$begingroup$
Thank you for this tenth solution. Is it not ironic how relative the lack of complexity this solution has, given the amount of time between its posting and the original problem?
$endgroup$
– Olive Stemforn
May 23 at 20:24
$begingroup$
Nice work,can;t believe I missed this.
$endgroup$
– hexomino
May 23 at 23:26
add a comment |
$begingroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.5k570139
answered May 22 at 21:15
PugmonkeyPugmonkey
3,9201221
$endgroup$
add a comment |
$begingroup$
$(sqrt3)^3! -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt3^3! -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8481935
$endgroup$
$begingroup$
Ah, yes, good spot!
$endgroup$
– hexomino
May 22 at 21:45
$begingroup$
@ Rupert Morrish - Good, your first one listed is a definitely newer type, because, for instance, if the square root symbol were to be dropped, the expression would have a different value.
$endgroup$
– Olive Stemforn
May 22 at 21:50
$begingroup$
@OliveStemforn: But the order of powers (which square root basically is) is unimportant (ie $a^b^c = a^c^b$) so are these two really different? To me it feels like it is the same principle as $3times3!+3!$ being different from $3!+3times3!$ - its exactly the same operations, just in a slightly different order...
$endgroup$
– Chris
May 23 at 15:15
$begingroup$
@chris The order of powers is definitely important. $2^3^5 ne 2^5^3$ The first is $2^3^5 = 2^243$. The second is $2^5^3=2^125$. Perhaps you mean $(a^b)^c = (a^c)^b$.
$endgroup$
– Trenin
May 23 at 17:34
$begingroup$
@Chris Because that is more applicable to this case anyways because $sqrt3^3!= (3^3!)^frac12=(3^frac12)^3!=(sqrt3)^3!$.
$endgroup$
– Trenin
May 23 at 17:50
|
show 1 more comment
$begingroup$
I feel like something should be able to be done with
$frac(3!)!30 = 24$
with judicious use of
decimal points
but I can't find it.
$frac(3!)!3^3 + 3 = 24$
uses 4 3s.
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8481935
$endgroup$
1
$begingroup$
:: : I will have to leave this computer in about five minutes from now (for up to as long as 20 hours later on) before I post for confirmations of remaining solutions.
$endgroup$
– Olive Stemforn
May 22 at 21:59
add a comment |
$begingroup$
Sorry not sure,
People will be working it out soon enough because voting is coming in fast...
answered 17 mins ago
user60673user60673
164
New contributor
user60673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
protected by Rubio♦ May 23 at 17:05
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).
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Ten solutions.
(i) $sqrt3^3! -3 = 24$
(ii) $left(frac3! + 3!3 right)! = 24$
(iii) $ left(frac3.3 - 3! right)! = 24$
(iv) $left(sqrtfrac3.3 + 3! right)! = 24$
(v) $left(3! - frac3!3 right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac33)! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt3)^3! -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8481935
answered May 22 at 21:29
hexominohexomino
52.2k4154246
$endgroup$
2
$begingroup$
@JonathanAllan That's what Olive's comment above was referring to.
$endgroup$
– hexomino
May 22 at 23:31
1
$begingroup$
@JonathanAllan the similar variant $(3 + fracsqrt3sqrt3)! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
4
$begingroup$
I'd say that i and ix are basically the same since the order of applying the square root and the power are irrelevant in the same way that 3x2 and 2x3 are the same (though that is just my subjective opinion).
$endgroup$
– Chris
May 23 at 14:45
1
$begingroup$
@Chris Yes I think you are right here, not sure if we really can consider them to be different considering what has been disallowed previously.
$endgroup$
– hexomino
May 23 at 15:32
2
$begingroup$
I see, and since it's more of a numerical "fluke" (in that it would only work for 3) that they cancel that stops them being classed as identical I guess.
$endgroup$
– Fifth_H0r5eman
May 23 at 15:33
|
show 19 more comments
$begingroup$
Ten solutions.
(i) $sqrt3^3! -3 = 24$
(ii) $left(frac3! + 3!3 right)! = 24$
(iii) $ left(frac3.3 - 3! right)! = 24$
(iv) $left(sqrtfrac3.3 + 3! right)! = 24$
(v) $left(3! - frac3!3 right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac33)! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt3)^3! -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8481935
answered May 22 at 21:29
hexominohexomino
52.2k4154246
$endgroup$
2
$begingroup$
@JonathanAllan That's what Olive's comment above was referring to.
$endgroup$
– hexomino
May 22 at 23:31
1
$begingroup$
@JonathanAllan the similar variant $(3 + fracsqrt3sqrt3)! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
4
$begingroup$
I'd say that i and ix are basically the same since the order of applying the square root and the power are irrelevant in the same way that 3x2 and 2x3 are the same (though that is just my subjective opinion).
$endgroup$
– Chris
May 23 at 14:45
1
$begingroup$
@Chris Yes I think you are right here, not sure if we really can consider them to be different considering what has been disallowed previously.
$endgroup$
– hexomino
May 23 at 15:32
2
$begingroup$
I see, and since it's more of a numerical "fluke" (in that it would only work for 3) that they cancel that stops them being classed as identical I guess.
$endgroup$
– Fifth_H0r5eman
May 23 at 15:33
|
show 19 more comments
$begingroup$
Ten solutions.
(i) $sqrt3^3! -3 = 24$
(ii) $left(frac3! + 3!3 right)! = 24$
(iii) $ left(frac3.3 - 3! right)! = 24$
(iv) $left(sqrtfrac3.3 + 3! right)! = 24$
(v) $left(3! - frac3!3 right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac33)! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt3)^3! -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8481935
answered May 22 at 21:29
hexominohexomino
52.2k4154246
$endgroup$
Ten solutions.
(i) $sqrt3^3! -3 = 24$
(ii) $left(frac3! + 3!3 right)! = 24$
(iii) $ left(frac3.3 - 3! right)! = 24$
(iv) $left(sqrtfrac3.3 + 3! right)! = 24$
(v) $left(3! - frac3!3 right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac33)! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt3)^3! -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8481935
answered May 22 at 21:29
hexominohexomino
52.2k4154246
edited May 23 at 20:26
Rupert Morrish
3,8481935
edited May 23 at 20:26
Rupert Morrish
3,8481935
edited May 23 at 20:26
Rupert Morrish
3,8481935
3,8481935
answered May 22 at 21:29
hexominohexomino
52.2k4154246
answered May 22 at 21:29
hexominohexomino
52.2k4154246
answered May 22 at 21:29
hexominohexomino
52.2k4154246
52.2k4154246
2
$begingroup$
@JonathanAllan That's what Olive's comment above was referring to.
$endgroup$
– hexomino
May 22 at 23:31
1
$begingroup$
@JonathanAllan the similar variant $(3 + fracsqrt3sqrt3)! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
4
$begingroup$
I'd say that i and ix are basically the same since the order of applying the square root and the power are irrelevant in the same way that 3x2 and 2x3 are the same (though that is just my subjective opinion).
$endgroup$
– Chris
May 23 at 14:45
1
$begingroup$
@Chris Yes I think you are right here, not sure if we really can consider them to be different considering what has been disallowed previously.
$endgroup$
– hexomino
May 23 at 15:32
2
$begingroup$
I see, and since it's more of a numerical "fluke" (in that it would only work for 3) that they cancel that stops them being classed as identical I guess.
$endgroup$
– Fifth_H0r5eman
May 23 at 15:33
|
show 19 more comments
2
$begingroup$
@JonathanAllan That's what Olive's comment above was referring to.
$endgroup$
– hexomino
May 22 at 23:31
1
$begingroup$
@JonathanAllan the similar variant $(3 + fracsqrt3sqrt3)! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
4
$begingroup$
I'd say that i and ix are basically the same since the order of applying the square root and the power are irrelevant in the same way that 3x2 and 2x3 are the same (though that is just my subjective opinion).
$endgroup$
– Chris
May 23 at 14:45
1
$begingroup$
@Chris Yes I think you are right here, not sure if we really can consider them to be different considering what has been disallowed previously.
$endgroup$
– hexomino
May 23 at 15:32
2
$begingroup$
I see, and since it's more of a numerical "fluke" (in that it would only work for 3) that they cancel that stops them being classed as identical I guess.
$endgroup$
– Fifth_H0r5eman
May 23 at 15:33
2
2
$begingroup$
@JonathanAllan That's what Olive's comment above was referring to.
$endgroup$
– hexomino
May 22 at 23:31
$begingroup$
@JonathanAllan That's what Olive's comment above was referring to.
$endgroup$
– hexomino
May 22 at 23:31
1
1
$begingroup$
@JonathanAllan the similar variant $(3 + fracsqrt3sqrt3)! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
$begingroup$
@JonathanAllan the similar variant $(3 + fracsqrt3sqrt3)! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
4
4
$begingroup$
I'd say that i and ix are basically the same since the order of applying the square root and the power are irrelevant in the same way that 3x2 and 2x3 are the same (though that is just my subjective opinion).
$endgroup$
– Chris
May 23 at 14:45
$begingroup$
I'd say that i and ix are basically the same since the order of applying the square root and the power are irrelevant in the same way that 3x2 and 2x3 are the same (though that is just my subjective opinion).
$endgroup$
– Chris
May 23 at 14:45
1
1
$begingroup$
@Chris Yes I think you are right here, not sure if we really can consider them to be different considering what has been disallowed previously.
$endgroup$
– hexomino
May 23 at 15:32
$begingroup$
@Chris Yes I think you are right here, not sure if we really can consider them to be different considering what has been disallowed previously.
$endgroup$
– hexomino
May 23 at 15:32
2
2
$begingroup$
I see, and since it's more of a numerical "fluke" (in that it would only work for 3) that they cancel that stops them being classed as identical I guess.
$endgroup$
– Fifth_H0r5eman
May 23 at 15:33
$begingroup$
I see, and since it's more of a numerical "fluke" (in that it would only work for 3) that they cancel that stops them being classed as identical I guess.
$endgroup$
– Fifth_H0r5eman
May 23 at 15:33
|
show 19 more comments
$begingroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
14.5k33084
answered May 22 at 21:05
Olive StemfornOlive Stemforn
733517
$endgroup$
16
$begingroup$
This solution came to mind after reading just the title of the question. I assumed it was the only one.
$endgroup$
– JollyJoker
May 23 at 7:32
add a comment |
$begingroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
14.5k33084
answered May 22 at 21:05
Olive StemfornOlive Stemforn
733517
$endgroup$
16
$begingroup$
This solution came to mind after reading just the title of the question. I assumed it was the only one.
$endgroup$
– JollyJoker
May 23 at 7:32
add a comment |
$begingroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
14.5k33084
answered May 22 at 21:05
Olive StemfornOlive Stemforn
733517
$endgroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
14.5k33084
answered May 22 at 21:05
Olive StemfornOlive Stemforn
733517
edited May 22 at 22:26
Dr Xorile
14.5k33084
edited May 22 at 22:26
Dr Xorile
14.5k33084
edited May 22 at 22:26
Dr Xorile
14.5k33084
14.5k33084
answered May 22 at 21:05
Olive StemfornOlive Stemforn
733517
answered May 22 at 21:05
Olive StemfornOlive Stemforn
733517
answered May 22 at 21:05
Olive StemfornOlive Stemforn
733517
733517
16
$begingroup$
This solution came to mind after reading just the title of the question. I assumed it was the only one.
$endgroup$
– JollyJoker
May 23 at 7:32
add a comment |
16
$begingroup$
This solution came to mind after reading just the title of the question. I assumed it was the only one.
$endgroup$
– JollyJoker
May 23 at 7:32
16
16
$begingroup$
This solution came to mind after reading just the title of the question. I assumed it was the only one.
$endgroup$
– JollyJoker
May 23 at 7:32
$begingroup$
This solution came to mind after reading just the title of the question. I assumed it was the only one.
$endgroup$
– JollyJoker
May 23 at 7:32
add a comment |
$begingroup$
Here is my first:
$(3 + frac33)! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
3,7231118
$endgroup$
1
$begingroup$
@ Weather Vane - Let us say that by adding the square root symbols, you have not changed the values of those. They are redundant. It's similar to writing 1! instead of 1 or 2! instead of 2. Although mathematically correct, I would not see them as "primitive solutions."
$endgroup$
– Olive Stemforn
May 22 at 21:44
1
$begingroup$
Ah OK, I was about to add another similar variant - removed.
$endgroup$
– Weather Vane
May 22 at 21:45
add a comment |
$begingroup$
Here is my first:
$(3 + frac33)! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
3,7231118
$endgroup$
1
$begingroup$
@ Weather Vane - Let us say that by adding the square root symbols, you have not changed the values of those. They are redundant. It's similar to writing 1! instead of 1 or 2! instead of 2. Although mathematically correct, I would not see them as "primitive solutions."
$endgroup$
– Olive Stemforn
May 22 at 21:44
1
$begingroup$
Ah OK, I was about to add another similar variant - removed.
$endgroup$
– Weather Vane
May 22 at 21:45
add a comment |
$begingroup$
Here is my first:
$(3 + frac33)! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
3,7231118
$endgroup$
Here is my first:
$(3 + frac33)! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
3,7231118
edited May 22 at 21:48
answered May 22 at 21:11
Weather VaneWeather Vane
3,7231118
answered May 22 at 21:11
Weather VaneWeather Vane
3,7231118
answered May 22 at 21:11
Weather VaneWeather Vane
3,7231118
3,7231118
1
$begingroup$
@ Weather Vane - Let us say that by adding the square root symbols, you have not changed the values of those. They are redundant. It's similar to writing 1! instead of 1 or 2! instead of 2. Although mathematically correct, I would not see them as "primitive solutions."
$endgroup$
– Olive Stemforn
May 22 at 21:44
1
$begingroup$
Ah OK, I was about to add another similar variant - removed.
$endgroup$
– Weather Vane
May 22 at 21:45
add a comment |
1
$begingroup$
@ Weather Vane - Let us say that by adding the square root symbols, you have not changed the values of those. They are redundant. It's similar to writing 1! instead of 1 or 2! instead of 2. Although mathematically correct, I would not see them as "primitive solutions."
$endgroup$
– Olive Stemforn
May 22 at 21:44
1
$begingroup$
Ah OK, I was about to add another similar variant - removed.
$endgroup$
– Weather Vane
May 22 at 21:45
1
1
$begingroup$
@ Weather Vane - Let us say that by adding the square root symbols, you have not changed the values of those. They are redundant. It's similar to writing 1! instead of 1 or 2! instead of 2. Although mathematically correct, I would not see them as "primitive solutions."
$endgroup$
– Olive Stemforn
May 22 at 21:44
$begingroup$
@ Weather Vane - Let us say that by adding the square root symbols, you have not changed the values of those. They are redundant. It's similar to writing 1! instead of 1 or 2! instead of 2. Although mathematically correct, I would not see them as "primitive solutions."
$endgroup$
– Olive Stemforn
May 22 at 21:44
1
1
$begingroup$
Ah OK, I was about to add another similar variant - removed.
$endgroup$
– Weather Vane
May 22 at 21:45
$begingroup$
Ah OK, I was about to add another similar variant - removed.
$endgroup$
– Weather Vane
May 22 at 21:45
add a comment |
$begingroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,0301646
$endgroup$
2
$begingroup$
Thank you for this tenth solution. Is it not ironic how relative the lack of complexity this solution has, given the amount of time between its posting and the original problem?
$endgroup$
– Olive Stemforn
May 23 at 20:24
$begingroup$
Nice work,can;t believe I missed this.
$endgroup$
– hexomino
May 23 at 23:26
add a comment |
$begingroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,0301646
$endgroup$
2
$begingroup$
Thank you for this tenth solution. Is it not ironic how relative the lack of complexity this solution has, given the amount of time between its posting and the original problem?
$endgroup$
– Olive Stemforn
May 23 at 20:24
$begingroup$
Nice work,can;t believe I missed this.
$endgroup$
– hexomino
May 23 at 23:26
add a comment |
$begingroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,0301646
$endgroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,0301646
answered May 23 at 19:36
TreninTrenin
8,0301646
answered May 23 at 19:36
TreninTrenin
8,0301646
answered May 23 at 19:36
TreninTrenin
8,0301646
8,0301646
2
$begingroup$
Thank you for this tenth solution. Is it not ironic how relative the lack of complexity this solution has, given the amount of time between its posting and the original problem?
$endgroup$
– Olive Stemforn
May 23 at 20:24
$begingroup$
Nice work,can;t believe I missed this.
$endgroup$
– hexomino
May 23 at 23:26
add a comment |
2
$begingroup$
Thank you for this tenth solution. Is it not ironic how relative the lack of complexity this solution has, given the amount of time between its posting and the original problem?
$endgroup$
– Olive Stemforn
May 23 at 20:24
$begingroup$
Nice work,can;t believe I missed this.
$endgroup$
– hexomino
May 23 at 23:26
2
2
$begingroup$
Thank you for this tenth solution. Is it not ironic how relative the lack of complexity this solution has, given the amount of time between its posting and the original problem?
$endgroup$
– Olive Stemforn
May 23 at 20:24
$begingroup$
Thank you for this tenth solution. Is it not ironic how relative the lack of complexity this solution has, given the amount of time between its posting and the original problem?
$endgroup$
– Olive Stemforn
May 23 at 20:24
$begingroup$
Nice work,can;t believe I missed this.
$endgroup$
– hexomino
May 23 at 23:26
$begingroup$
Nice work,can;t believe I missed this.
$endgroup$
– hexomino
May 23 at 23:26
add a comment |
$begingroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.5k570139
answered May 22 at 21:15
PugmonkeyPugmonkey
3,9201221
$endgroup$
add a comment |
$begingroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.5k570139
answered May 22 at 21:15
PugmonkeyPugmonkey
3,9201221
$endgroup$
add a comment |
$begingroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.5k570139
answered May 22 at 21:15
PugmonkeyPugmonkey
3,9201221
$endgroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.5k570139
answered May 22 at 21:15
PugmonkeyPugmonkey
3,9201221
edited May 23 at 14:31
GentlePurpleRain♦
17.5k570139
edited May 23 at 14:31
GentlePurpleRain♦
17.5k570139
edited May 23 at 14:31
GentlePurpleRain♦
17.5k570139
17.5k570139
answered May 22 at 21:15
PugmonkeyPugmonkey
3,9201221
answered May 22 at 21:15
PugmonkeyPugmonkey
3,9201221
answered May 22 at 21:15
PugmonkeyPugmonkey
3,9201221
3,9201221
add a comment |
add a comment |
$begingroup$
$(sqrt3)^3! -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt3^3! -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8481935
$endgroup$
$begingroup$
Ah, yes, good spot!
$endgroup$
– hexomino
May 22 at 21:45
$begingroup$
@ Rupert Morrish - Good, your first one listed is a definitely newer type, because, for instance, if the square root symbol were to be dropped, the expression would have a different value.
$endgroup$
– Olive Stemforn
May 22 at 21:50
$begingroup$
@OliveStemforn: But the order of powers (which square root basically is) is unimportant (ie $a^b^c = a^c^b$) so are these two really different? To me it feels like it is the same principle as $3times3!+3!$ being different from $3!+3times3!$ - its exactly the same operations, just in a slightly different order...
$endgroup$
– Chris
May 23 at 15:15
$begingroup$
@chris The order of powers is definitely important. $2^3^5 ne 2^5^3$ The first is $2^3^5 = 2^243$. The second is $2^5^3=2^125$. Perhaps you mean $(a^b)^c = (a^c)^b$.
$endgroup$
– Trenin
May 23 at 17:34
$begingroup$
@Chris Because that is more applicable to this case anyways because $sqrt3^3!= (3^3!)^frac12=(3^frac12)^3!=(sqrt3)^3!$.
$endgroup$
– Trenin
May 23 at 17:50
|
show 1 more comment
$begingroup$
$(sqrt3)^3! -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt3^3! -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8481935
$endgroup$
$begingroup$
Ah, yes, good spot!
$endgroup$
– hexomino
May 22 at 21:45
$begingroup$
@ Rupert Morrish - Good, your first one listed is a definitely newer type, because, for instance, if the square root symbol were to be dropped, the expression would have a different value.
$endgroup$
– Olive Stemforn
May 22 at 21:50
$begingroup$
@OliveStemforn: But the order of powers (which square root basically is) is unimportant (ie $a^b^c = a^c^b$) so are these two really different? To me it feels like it is the same principle as $3times3!+3!$ being different from $3!+3times3!$ - its exactly the same operations, just in a slightly different order...
$endgroup$
– Chris
May 23 at 15:15
$begingroup$
@chris The order of powers is definitely important. $2^3^5 ne 2^5^3$ The first is $2^3^5 = 2^243$. The second is $2^5^3=2^125$. Perhaps you mean $(a^b)^c = (a^c)^b$.
$endgroup$
– Trenin
May 23 at 17:34
$begingroup$
@Chris Because that is more applicable to this case anyways because $sqrt3^3!= (3^3!)^frac12=(3^frac12)^3!=(sqrt3)^3!$.
$endgroup$
– Trenin
May 23 at 17:50
|
show 1 more comment
$begingroup$
$(sqrt3)^3! -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt3^3! -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8481935
$endgroup$
$(sqrt3)^3! -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt3^3! -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8481935
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8481935
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8481935
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8481935
3,8481935
$begingroup$
Ah, yes, good spot!
$endgroup$
– hexomino
May 22 at 21:45
$begingroup$
@ Rupert Morrish - Good, your first one listed is a definitely newer type, because, for instance, if the square root symbol were to be dropped, the expression would have a different value.
$endgroup$
– Olive Stemforn
May 22 at 21:50
$begingroup$
@OliveStemforn: But the order of powers (which square root basically is) is unimportant (ie $a^b^c = a^c^b$) so are these two really different? To me it feels like it is the same principle as $3times3!+3!$ being different from $3!+3times3!$ - its exactly the same operations, just in a slightly different order...
$endgroup$
– Chris
May 23 at 15:15
$begingroup$
@chris The order of powers is definitely important. $2^3^5 ne 2^5^3$ The first is $2^3^5 = 2^243$. The second is $2^5^3=2^125$. Perhaps you mean $(a^b)^c = (a^c)^b$.
$endgroup$
– Trenin
May 23 at 17:34
$begingroup$
@Chris Because that is more applicable to this case anyways because $sqrt3^3!= (3^3!)^frac12=(3^frac12)^3!=(sqrt3)^3!$.
$endgroup$
– Trenin
May 23 at 17:50
|
show 1 more comment
$begingroup$
Ah, yes, good spot!
$endgroup$
– hexomino
May 22 at 21:45
$begingroup$
@ Rupert Morrish - Good, your first one listed is a definitely newer type, because, for instance, if the square root symbol were to be dropped, the expression would have a different value.
$endgroup$
– Olive Stemforn
May 22 at 21:50
$begingroup$
@OliveStemforn: But the order of powers (which square root basically is) is unimportant (ie $a^b^c = a^c^b$) so are these two really different? To me it feels like it is the same principle as $3times3!+3!$ being different from $3!+3times3!$ - its exactly the same operations, just in a slightly different order...
$endgroup$
– Chris
May 23 at 15:15
$begingroup$
@chris The order of powers is definitely important. $2^3^5 ne 2^5^3$ The first is $2^3^5 = 2^243$. The second is $2^5^3=2^125$. Perhaps you mean $(a^b)^c = (a^c)^b$.
$endgroup$
– Trenin
May 23 at 17:34
$begingroup$
@Chris Because that is more applicable to this case anyways because $sqrt3^3!= (3^3!)^frac12=(3^frac12)^3!=(sqrt3)^3!$.
$endgroup$
– Trenin
May 23 at 17:50
$begingroup$
Ah, yes, good spot!
$endgroup$
– hexomino
May 22 at 21:45
$begingroup$
Ah, yes, good spot!
$endgroup$
– hexomino
May 22 at 21:45
$begingroup$
@ Rupert Morrish - Good, your first one listed is a definitely newer type, because, for instance, if the square root symbol were to be dropped, the expression would have a different value.
$endgroup$
– Olive Stemforn
May 22 at 21:50
$begingroup$
@ Rupert Morrish - Good, your first one listed is a definitely newer type, because, for instance, if the square root symbol were to be dropped, the expression would have a different value.
$endgroup$
– Olive Stemforn
May 22 at 21:50
$begingroup$
@OliveStemforn: But the order of powers (which square root basically is) is unimportant (ie $a^b^c = a^c^b$) so are these two really different? To me it feels like it is the same principle as $3times3!+3!$ being different from $3!+3times3!$ - its exactly the same operations, just in a slightly different order...
$endgroup$
– Chris
May 23 at 15:15
$begingroup$
@OliveStemforn: But the order of powers (which square root basically is) is unimportant (ie $a^b^c = a^c^b$) so are these two really different? To me it feels like it is the same principle as $3times3!+3!$ being different from $3!+3times3!$ - its exactly the same operations, just in a slightly different order...
$endgroup$
– Chris
May 23 at 15:15
$begingroup$
@chris The order of powers is definitely important. $2^3^5 ne 2^5^3$ The first is $2^3^5 = 2^243$. The second is $2^5^3=2^125$. Perhaps you mean $(a^b)^c = (a^c)^b$.
$endgroup$
– Trenin
May 23 at 17:34
$begingroup$
@chris The order of powers is definitely important. $2^3^5 ne 2^5^3$ The first is $2^3^5 = 2^243$. The second is $2^5^3=2^125$. Perhaps you mean $(a^b)^c = (a^c)^b$.
$endgroup$
– Trenin
May 23 at 17:34
$begingroup$
@Chris Because that is more applicable to this case anyways because $sqrt3^3!= (3^3!)^frac12=(3^frac12)^3!=(sqrt3)^3!$.
$endgroup$
– Trenin
May 23 at 17:50
$begingroup$
@Chris Because that is more applicable to this case anyways because $sqrt3^3!= (3^3!)^frac12=(3^frac12)^3!=(sqrt3)^3!$.
$endgroup$
– Trenin
May 23 at 17:50
|
show 1 more comment
$begingroup$
I feel like something should be able to be done with
$frac(3!)!30 = 24$
with judicious use of
decimal points
but I can't find it.
$frac(3!)!3^3 + 3 = 24$
uses 4 3s.
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8481935
$endgroup$
1
$begingroup$
:: : I will have to leave this computer in about five minutes from now (for up to as long as 20 hours later on) before I post for confirmations of remaining solutions.
$endgroup$
– Olive Stemforn
May 22 at 21:59
add a comment |
$begingroup$
I feel like something should be able to be done with
$frac(3!)!30 = 24$
with judicious use of
decimal points
but I can't find it.
$frac(3!)!3^3 + 3 = 24$
uses 4 3s.
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8481935
$endgroup$
1
$begingroup$
:: : I will have to leave this computer in about five minutes from now (for up to as long as 20 hours later on) before I post for confirmations of remaining solutions.
$endgroup$
– Olive Stemforn
May 22 at 21:59
add a comment |
$begingroup$
I feel like something should be able to be done with
$frac(3!)!30 = 24$
with judicious use of
decimal points
but I can't find it.
$frac(3!)!3^3 + 3 = 24$
uses 4 3s.
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8481935
$endgroup$
I feel like something should be able to be done with
$frac(3!)!30 = 24$
with judicious use of
decimal points
but I can't find it.
$frac(3!)!3^3 + 3 = 24$
uses 4 3s.
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8481935
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8481935
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8481935
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8481935
3,8481935
1
$begingroup$
:: : I will have to leave this computer in about five minutes from now (for up to as long as 20 hours later on) before I post for confirmations of remaining solutions.
$endgroup$
– Olive Stemforn
May 22 at 21:59
add a comment |
1
$begingroup$
:: : I will have to leave this computer in about five minutes from now (for up to as long as 20 hours later on) before I post for confirmations of remaining solutions.
$endgroup$
– Olive Stemforn
May 22 at 21:59
1
1
$begingroup$
:: : I will have to leave this computer in about five minutes from now (for up to as long as 20 hours later on) before I post for confirmations of remaining solutions.
$endgroup$
– Olive Stemforn
May 22 at 21:59
$begingroup$
:: : I will have to leave this computer in about five minutes from now (for up to as long as 20 hours later on) before I post for confirmations of remaining solutions.
$endgroup$
– Olive Stemforn
May 22 at 21:59
add a comment |
$begingroup$
Sorry not sure,
People will be working it out soon enough because voting is coming in fast...
answered 17 mins ago
user60673user60673
164
New contributor
user60673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Sorry not sure,
People will be working it out soon enough because voting is coming in fast...
answered 17 mins ago
user60673user60673
164
New contributor
user60673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Sorry not sure,
People will be working it out soon enough because voting is coming in fast...
answered 17 mins ago
user60673user60673
164
New contributor
user60673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Sorry not sure,
People will be working it out soon enough because voting is coming in fast...
answered 17 mins ago
user60673user60673
164
New contributor
user60673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 17 mins ago
user60673user60673
164
New contributor
user60673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 17 mins ago
user60673user60673
164
answered 17 mins ago
user60673user60673
164
164
New contributor
user60673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user60673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
protected by Rubio♦ May 23 at 17:05
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?
6
$begingroup$
Let's please not make "non-serious" or otherwise very clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers.
$endgroup$
– Rubio♦
May 23 at 14:34
$begingroup$
Has a correct answer been given? If so, please don't forget to $colorgreencheckmark smalltextAccept$ it :)
$endgroup$
– Rubio♦
May 28 at 4:30