Make 24 using exactly three 3s
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15
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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
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add a comment |
15
$begingroup$
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
7655 silver badges17 bronze badges
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6
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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.
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– Rubio♦
May 23 at 14:34
add a comment |
15
15
15
4
$begingroup$
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
7655 silver badges17 bronze badges
$endgroup$
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
7655 silver badges17 bronze badges
asked May 22 at 21:03
Olive StemfornOlive Stemforn
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edited May 23 at 19:25
Olive Stemforn
asked May 22 at 21:03
Olive StemfornOlive Stemforn
7655 silver badges17 bronze badges
asked May 22 at 21:03
Olive StemfornOlive Stemforn
7655 silver badges17 bronze badges
asked May 22 at 21:03
Olive StemfornOlive Stemforn
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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
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
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
add a comment |
7 Answers
7
active
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36
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Ten solutions.
(i) $sqrt{3^{3!}} -3 = 24$
(ii) $left(frac{3! + 3!}{3} right)! = 24$
(iii) $ left(frac{3}{.3} - 3! right)! = 24$
(iv) $left(sqrt{frac{3}{.3} + 3!} right)! = 24$
(v) $left(3! - frac{3!}{3} right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac{3}{3})! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt{3})^{3!} -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
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answered May 22 at 21:29
hexominohexomino
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2
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@JonathanAllan That's what Olive's comment above was referring to.
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– hexomino
May 22 at 23:31
1
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@JonathanAllan the similar variant $(3 + frac{sqrt{3}}{sqrt{3}})! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
4
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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).
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– Chris
May 23 at 14:45
1
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@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.
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– hexomino
May 23 at 15:32
2
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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.
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– Fifth_H0r5eman
May 23 at 15:33
|
show 19 more comments
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(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
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answered May 22 at 21:05
Olive StemfornOlive Stemforn
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16
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This solution came to mind after reading just the title of the question. I assumed it was the only one.
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– JollyJoker
May 23 at 7:32
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13
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Here is my first:
$(3 + frac{3}{3})! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
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1
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@ 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."
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– Olive Stemforn
May 22 at 21:44
1
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Ah OK, I was about to add another similar variant - removed.
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– Weather Vane
May 22 at 21:45
add a comment |
11
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Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
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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?
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– Olive Stemforn
May 23 at 20:24
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Nice work,can;t believe I missed this.
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– hexomino
May 23 at 23:26
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9
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Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
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answered May 22 at 21:15
PugmonkeyPugmonkey
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8
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$(sqrt{3})^{3!} -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt{3^{3!}} -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
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Ah, yes, good spot!
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– hexomino
May 22 at 21:45
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@ 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.
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– Olive Stemforn
May 22 at 21:50
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@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...
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– Chris
May 23 at 15:15
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@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 $(a^b)^c = (a^c)^b$.
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– Trenin
May 23 at 17:34
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@Chris Because that is more applicable to this case anyways because $sqrt{3^{3!}}= (3^{3!})^{frac{1}{2}}=(3^{frac{1}{2}})^{3!}=(sqrt{3})^{3!}$.
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– Trenin
May 23 at 17:50
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show 1 more comment
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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
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1
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:: : 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.
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– Olive Stemforn
May 22 at 21:59
add a comment |
protected by Rubio♦ May 23 at 17:05
Thank you for your interest in this question.
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
36
$begingroup$
Ten solutions.
(i) $sqrt{3^{3!}} -3 = 24$
(ii) $left(frac{3! + 3!}{3} right)! = 24$
(iii) $ left(frac{3}{.3} - 3! right)! = 24$
(iv) $left(sqrt{frac{3}{.3} + 3!} right)! = 24$
(v) $left(3! - frac{3!}{3} right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac{3}{3})! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt{3})^{3!} -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:29
hexominohexomino
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2
$begingroup$
@JonathanAllan That's what Olive's comment above was referring to.
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– hexomino
May 22 at 23:31
1
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@JonathanAllan the similar variant $(3 + frac{sqrt{3}}{sqrt{3}})! = 24$ 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
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@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
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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
36
$begingroup$
Ten solutions.
(i) $sqrt{3^{3!}} -3 = 24$
(ii) $left(frac{3! + 3!}{3} right)! = 24$
(iii) $ left(frac{3}{.3} - 3! right)! = 24$
(iv) $left(sqrt{frac{3}{.3} + 3!} right)! = 24$
(v) $left(3! - frac{3!}{3} right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac{3}{3})! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt{3})^{3!} -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:29
hexominohexomino
58.4k5 gold badges169 silver badges267 bronze badges
$endgroup$
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 $(3 + frac{sqrt{3}}{sqrt{3}})! = 24$ 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
36
36
36
$begingroup$
Ten solutions.
(i) $sqrt{3^{3!}} -3 = 24$
(ii) $left(frac{3! + 3!}{3} right)! = 24$
(iii) $ left(frac{3}{.3} - 3! right)! = 24$
(iv) $left(sqrt{frac{3}{.3} + 3!} right)! = 24$
(v) $left(3! - frac{3!}{3} right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac{3}{3})! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt{3})^{3!} -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:29
hexominohexomino
58.4k5 gold badges169 silver badges267 bronze badges
$endgroup$
Ten solutions.
(i) $sqrt{3^{3!}} -3 = 24$
(ii) $left(frac{3! + 3!}{3} right)! = 24$
(iii) $ left(frac{3}{.3} - 3! right)! = 24$
(iv) $left(sqrt{frac{3}{.3} + 3!} right)! = 24$
(v) $left(3! - frac{3!}{3} right)! = 24$
(vi) $(3times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + frac{3}{3})! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(sqrt{3})^{3!} -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)
edited May 23 at 20:26
Rupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:29
hexominohexomino
58.4k5 gold badges169 silver badges267 bronze badges
edited May 23 at 20:26
Rupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
edited May 23 at 20:26
Rupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
edited May 23 at 20:26
Rupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:29
hexominohexomino
58.4k5 gold badges169 silver badges267 bronze badges
answered May 22 at 21:29
hexominohexomino
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answered May 22 at 21:29
hexominohexomino
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58.4k5 gold badges169 silver badges267 bronze badges
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 $(3 + frac{sqrt{3}}{sqrt{3}})! = 24$ 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 $(3 + frac{sqrt{3}}{sqrt{3}})! = 24$ 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 $(3 + frac{sqrt{3}}{sqrt{3}})! = 24$ was banned too.
$endgroup$
– Weather Vane
May 22 at 23:54
$begingroup$
@JonathanAllan the similar variant $(3 + frac{sqrt{3}}{sqrt{3}})! = 24$ 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
20
$begingroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
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answered May 22 at 21:05
Olive StemfornOlive Stemforn
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16
$begingroup$
This solution came to mind after reading just the title of the question. I assumed it was the only one.
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– JollyJoker
May 23 at 7:32
add a comment |
20
$begingroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
14.7k3 gold badges31 silver badges89 bronze badges
answered May 22 at 21:05
Olive StemfornOlive Stemforn
7655 silver badges17 bronze badges
$endgroup$
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 |
20
20
20
$begingroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
14.7k3 gold badges31 silver badges89 bronze badges
answered May 22 at 21:05
Olive StemfornOlive Stemforn
7655 silver badges17 bronze badges
$endgroup$
(edit) I will post one of the 10 solutions as an example:
$3^3 - 3 = 24$
edited May 22 at 22:26
Dr Xorile
14.7k3 gold badges31 silver badges89 bronze badges
answered May 22 at 21:05
Olive StemfornOlive Stemforn
7655 silver badges17 bronze badges
edited May 22 at 22:26
Dr Xorile
14.7k3 gold badges31 silver badges89 bronze badges
edited May 22 at 22:26
Dr Xorile
14.7k3 gold badges31 silver badges89 bronze badges
edited May 22 at 22:26
Dr Xorile
14.7k3 gold badges31 silver badges89 bronze badges
14.7k3 gold badges31 silver badges89 bronze badges
answered May 22 at 21:05
Olive StemfornOlive Stemforn
7655 silver badges17 bronze badges
answered May 22 at 21:05
Olive StemfornOlive Stemforn
7655 silver badges17 bronze badges
answered May 22 at 21:05
Olive StemfornOlive Stemforn
7655 silver badges17 bronze badges
7655 silver badges17 bronze badges
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 |
13
$begingroup$
Here is my first:
$(3 + frac{3}{3})! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
5,3461 gold badge3 silver badges23 bronze badges
$endgroup$
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 |
13
$begingroup$
Here is my first:
$(3 + frac{3}{3})! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
5,3461 gold badge3 silver badges23 bronze badges
$endgroup$
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 |
13
13
13
$begingroup$
Here is my first:
$(3 + frac{3}{3})! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
5,3461 gold badge3 silver badges23 bronze badges
$endgroup$
Here is my first:
$(3 + frac{3}{3})! = 24$
answered May 22 at 21:11
Weather VaneWeather Vane
5,3461 gold badge3 silver badges23 bronze badges
edited May 22 at 21:48
answered May 22 at 21:11
Weather VaneWeather Vane
5,3461 gold badge3 silver badges23 bronze badges
answered May 22 at 21:11
Weather VaneWeather Vane
5,3461 gold badge3 silver badges23 bronze badges
answered May 22 at 21:11
Weather VaneWeather Vane
5,3461 gold badge3 silver badges23 bronze badges
5,3461 gold badge3 silver badges23 bronze badges
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 |
11
$begingroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,35817 silver badges49 bronze badges
$endgroup$
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 |
11
$begingroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,35817 silver badges49 bronze badges
$endgroup$
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 |
11
11
11
$begingroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,35817 silver badges49 bronze badges
$endgroup$
Tenth solution from hint
$$left(3+(3-3)!right)!$$
answered May 23 at 19:36
TreninTrenin
8,35817 silver badges49 bronze badges
answered May 23 at 19:36
TreninTrenin
8,35817 silver badges49 bronze badges
answered May 23 at 19:36
TreninTrenin
8,35817 silver badges49 bronze badges
answered May 23 at 19:36
TreninTrenin
8,35817 silver badges49 bronze badges
8,35817 silver badges49 bronze badges
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 |
9
$begingroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.9k6 gold badges71 silver badges141 bronze badges
answered May 22 at 21:15
PugmonkeyPugmonkey
4,05512 silver badges22 bronze badges
$endgroup$
add a comment |
9
$begingroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.9k6 gold badges71 silver badges141 bronze badges
answered May 22 at 21:15
PugmonkeyPugmonkey
4,05512 silver badges22 bronze badges
$endgroup$
add a comment |
9
9
9
$begingroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.9k6 gold badges71 silver badges141 bronze badges
answered May 22 at 21:15
PugmonkeyPugmonkey
4,05512 silver badges22 bronze badges
$endgroup$
Here is one possible solution:
$3times3!+3!$
edited May 23 at 14:31
GentlePurpleRain♦
17.9k6 gold badges71 silver badges141 bronze badges
answered May 22 at 21:15
PugmonkeyPugmonkey
4,05512 silver badges22 bronze badges
edited May 23 at 14:31
GentlePurpleRain♦
17.9k6 gold badges71 silver badges141 bronze badges
edited May 23 at 14:31
GentlePurpleRain♦
17.9k6 gold badges71 silver badges141 bronze badges
edited May 23 at 14:31
GentlePurpleRain♦
17.9k6 gold badges71 silver badges141 bronze badges
17.9k6 gold badges71 silver badges141 bronze badges
answered May 22 at 21:15
PugmonkeyPugmonkey
4,05512 silver badges22 bronze badges
answered May 22 at 21:15
PugmonkeyPugmonkey
4,05512 silver badges22 bronze badges
answered May 22 at 21:15
PugmonkeyPugmonkey
4,05512 silver badges22 bronze badges
4,05512 silver badges22 bronze badges
add a comment |
add a comment |
8
$begingroup$
$(sqrt{3})^{3!} -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt{3^{3!}} -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
$endgroup$
$begingroup$
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 $sqrt{3^{3!}}= (3^{3!})^{frac{1}{2}}=(3^{frac{1}{2}})^{3!}=(sqrt{3})^{3!}$.
$endgroup$
– Trenin
May 23 at 17:50
|
show 1 more comment
8
$begingroup$
$(sqrt{3})^{3!} -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt{3^{3!}} -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
$endgroup$
$begingroup$
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 $sqrt{3^{3!}}= (3^{3!})^{frac{1}{2}}=(3^{frac{1}{2}})^{3!}=(sqrt{3})^{3!}$.
$endgroup$
– Trenin
May 23 at 17:50
|
show 1 more comment
8
8
8
$begingroup$
$(sqrt{3})^{3!} -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt{3^{3!}} -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
$endgroup$
$(sqrt{3})^{3!} -3 = 24$
Which is obviously only slightly different to hexomino's:
(i) $sqrt{3^{3!}} -3 = 24$
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:39
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
3,8581 gold badge10 silver badges35 bronze badges
$begingroup$
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 $sqrt{3^{3!}}= (3^{3!})^{frac{1}{2}}=(3^{frac{1}{2}})^{3!}=(sqrt{3})^{3!}$.
$endgroup$
– 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 $sqrt{3^{3!}}= (3^{3!})^{frac{1}{2}}=(3^{frac{1}{2}})^{3!}=(sqrt{3})^{3!}$.
$endgroup$
– 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 $sqrt{3^{3!}}= (3^{3!})^{frac{1}{2}}=(3^{frac{1}{2}})^{3!}=(sqrt{3})^{3!}$.
$endgroup$
– Trenin
May 23 at 17:50
$begingroup$
@Chris Because that is more applicable to this case anyways because $sqrt{3^{3!}}= (3^{3!})^{frac{1}{2}}=(3^{frac{1}{2}})^{3!}=(sqrt{3})^{3!}$.
$endgroup$
– Trenin
May 23 at 17:50
|
show 1 more comment
3
$begingroup$
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
3,8581 gold badge10 silver badges35 bronze badges
$endgroup$
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 |
3
$begingroup$
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
3,8581 gold badge10 silver badges35 bronze badges
$endgroup$
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 |
3
3
3
$begingroup$
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
3,8581 gold badge10 silver badges35 bronze badges
$endgroup$
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
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
answered May 22 at 21:50
Rupert MorrishRupert Morrish
3,8581 gold badge10 silver badges35 bronze badges
3,8581 gold badge10 silver badges35 bronze badges
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 |
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