Mixing
Use 6, 5 and 3 to make 57
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Assemble a formula using the numbers $5$, $6$, and $3$ in any order to make $57$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $3$, $6$, or $5$. Operands may of course also be derived from calculations e.g. $3+(6*5)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $6$ and $5$ to make the number $65$) if you wish. You may only use each of the starting digits once and you must use all three of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $57$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $5!! = 5 times 3 times 1$ are not allowed, but factorials of factorials are fine, such as $((3)!)! = 6!$. I will upvote answers with double, triple and n-druple-factorials which get $57$, but will not mark them as correct.
A finite number of operations should be used, but answers with an infinite number of operations will get +1 from me...
mathematics formation-of-numbers
asked 38 mins ago
tom
1,8501325
add a comment |Â
up vote
1
down vote
favorite
Assemble a formula using the numbers $5$, $6$, and $3$ in any order to make $57$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $3$, $6$, or $5$. Operands may of course also be derived from calculations e.g. $3+(6*5)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $6$ and $5$ to make the number $65$) if you wish. You may only use each of the starting digits once and you must use all three of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $57$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $5!! = 5 times 3 times 1$ are not allowed, but factorials of factorials are fine, such as $((3)!)! = 6!$. I will upvote answers with double, triple and n-druple-factorials which get $57$, but will not mark them as correct.
A finite number of operations should be used, but answers with an infinite number of operations will get +1 from me...
mathematics formation-of-numbers
asked 38 mins ago
tom
1,8501325
1
Tom, you're posting a lot of these puzzles and they're all basically the same and -- excuse my frankness! -- becoming rather boring. Maybe give them a rest for a while?
– Gareth McCaughan♦
25 mins ago
@GarethMcCaughan to be honest, some of them are very original, requires good logic deduction. but yes, would be good to give a rest for a while :)
– Oray
23 mins ago
@GarethMcCaughan you could say the same about the large number of Riley riddles. Some people like one kind of puzzle, but not the other, etc. Isn't the time to stop when no-one answers, or there are a lot of very quick correct answers?
– Weather Vane
19 mins ago
@GarethMcCaughan - point taken... I will retire from these riddles.
– tom
4 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Assemble a formula using the numbers $5$, $6$, and $3$ in any order to make $57$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $3$, $6$, or $5$. Operands may of course also be derived from calculations e.g. $3+(6*5)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $6$ and $5$ to make the number $65$) if you wish. You may only use each of the starting digits once and you must use all three of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $57$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $5!! = 5 times 3 times 1$ are not allowed, but factorials of factorials are fine, such as $((3)!)! = 6!$. I will upvote answers with double, triple and n-druple-factorials which get $57$, but will not mark them as correct.
A finite number of operations should be used, but answers with an infinite number of operations will get +1 from me...
mathematics formation-of-numbers
asked 38 mins ago
tom
1,8501325
Assemble a formula using the numbers $5$, $6$, and $3$ in any order to make $57$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $3$, $6$, or $5$. Operands may of course also be derived from calculations e.g. $3+(6*5)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $6$ and $5$ to make the number $65$) if you wish. You may only use each of the starting digits once and you must use all three of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $57$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $5!! = 5 times 3 times 1$ are not allowed, but factorials of factorials are fine, such as $((3)!)! = 6!$. I will upvote answers with double, triple and n-druple-factorials which get $57$, but will not mark them as correct.
A finite number of operations should be used, but answers with an infinite number of operations will get +1 from me...
mathematics formation-of-numbers
mathematics formation-of-numbers
asked 38 mins ago
tom
1,8501325
asked 38 mins ago
tom
1,8501325
edited 29 mins ago
asked 38 mins ago
tom
1,8501325
asked 38 mins ago
tom
1,8501325
asked 38 mins ago
tom
1,8501325
1,8501325
1
Tom, you're posting a lot of these puzzles and they're all basically the same and -- excuse my frankness! -- becoming rather boring. Maybe give them a rest for a while?
– Gareth McCaughan♦
25 mins ago
@GarethMcCaughan to be honest, some of them are very original, requires good logic deduction. but yes, would be good to give a rest for a while :)
– Oray
23 mins ago
@GarethMcCaughan you could say the same about the large number of Riley riddles. Some people like one kind of puzzle, but not the other, etc. Isn't the time to stop when no-one answers, or there are a lot of very quick correct answers?
– Weather Vane
19 mins ago
@GarethMcCaughan - point taken... I will retire from these riddles.
– tom
4 mins ago
add a comment |Â
1
Tom, you're posting a lot of these puzzles and they're all basically the same and -- excuse my frankness! -- becoming rather boring. Maybe give them a rest for a while?
– Gareth McCaughan♦
25 mins ago
@GarethMcCaughan to be honest, some of them are very original, requires good logic deduction. but yes, would be good to give a rest for a while :)
– Oray
23 mins ago
@GarethMcCaughan you could say the same about the large number of Riley riddles. Some people like one kind of puzzle, but not the other, etc. Isn't the time to stop when no-one answers, or there are a lot of very quick correct answers?
– Weather Vane
19 mins ago
@GarethMcCaughan - point taken... I will retire from these riddles.
– tom
4 mins ago
1
1
Tom, you're posting a lot of these puzzles and they're all basically the same and -- excuse my frankness! -- becoming rather boring. Maybe give them a rest for a while?
– Gareth McCaughan♦
25 mins ago
Tom, you're posting a lot of these puzzles and they're all basically the same and -- excuse my frankness! -- becoming rather boring. Maybe give them a rest for a while?
– Gareth McCaughan♦
25 mins ago
@GarethMcCaughan to be honest, some of them are very original, requires good logic deduction. but yes, would be good to give a rest for a while :)
– Oray
23 mins ago
@GarethMcCaughan to be honest, some of them are very original, requires good logic deduction. but yes, would be good to give a rest for a while :)
– Oray
23 mins ago
@GarethMcCaughan you could say the same about the large number of Riley riddles. Some people like one kind of puzzle, but not the other, etc. Isn't the time to stop when no-one answers, or there are a lot of very quick correct answers?
– Weather Vane
19 mins ago
@GarethMcCaughan you could say the same about the large number of Riley riddles. Some people like one kind of puzzle, but not the other, etc. Isn't the time to stop when no-one answers, or there are a lot of very quick correct answers?
– Weather Vane
19 mins ago
@GarethMcCaughan - point taken... I will retire from these riddles.
– tom
4 mins ago
@GarethMcCaughan - point taken... I will retire from these riddles.
– tom
4 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Possible answer
$5! - 63 = 57$
answered 19 mins ago
hexomino
30.7k293147
Great job well done
– tom
6 mins ago
add a comment |Â
up vote
3
down vote
hexomino beat me, but I have another:
$sqrt5times6!-3=57$
answered 19 mins ago
LegionMammal978
947512
IMO this answer without number concatenation is better.
– Weather Vane
13 mins ago
@WeatherVane, but both are within the rules, so the sightly earlier answer gets the solution, but both get plus one.
– tom
5 mins ago
add a comment |Â
up vote
1
down vote
Can I have:
$56+sqrtsqrtdots3$
answered 21 mins ago
JonMark Perry
14.9k52972
That's a dotty answer. Interesting but is it in the rules?
– Weather Vane
14 mins ago
@WeatherVane it counts as an infinite number of operations, so gets plus one.
– tom
5 mins ago
@tom I see you changed the rules.
– Weather Vane
12 secs ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Possible answer
$5! - 63 = 57$
answered 19 mins ago
hexomino
30.7k293147
Great job well done
– tom
6 mins ago
add a comment |Â
up vote
3
down vote
accepted
Possible answer
$5! - 63 = 57$
answered 19 mins ago
hexomino
30.7k293147
Great job well done
– tom
6 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Possible answer
$5! - 63 = 57$
answered 19 mins ago
hexomino
30.7k293147
Possible answer
$5! - 63 = 57$
answered 19 mins ago
hexomino
30.7k293147
answered 19 mins ago
hexomino
30.7k293147
answered 19 mins ago
hexomino
30.7k293147
answered 19 mins ago
hexomino
30.7k293147
30.7k293147
Great job well done
– tom
6 mins ago
add a comment |Â
Great job well done
– tom
6 mins ago
Great job well done
– tom
6 mins ago
Great job well done
– tom
6 mins ago
add a comment |Â
up vote
3
down vote
hexomino beat me, but I have another:
$sqrt5times6!-3=57$
answered 19 mins ago
LegionMammal978
947512
IMO this answer without number concatenation is better.
– Weather Vane
13 mins ago
@WeatherVane, but both are within the rules, so the sightly earlier answer gets the solution, but both get plus one.
– tom
5 mins ago
add a comment |Â
up vote
3
down vote
hexomino beat me, but I have another:
$sqrt5times6!-3=57$
answered 19 mins ago
LegionMammal978
947512
IMO this answer without number concatenation is better.
– Weather Vane
13 mins ago
@WeatherVane, but both are within the rules, so the sightly earlier answer gets the solution, but both get plus one.
– tom
5 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
hexomino beat me, but I have another:
$sqrt5times6!-3=57$
answered 19 mins ago
LegionMammal978
947512
hexomino beat me, but I have another:
$sqrt5times6!-3=57$
answered 19 mins ago
LegionMammal978
947512
answered 19 mins ago
LegionMammal978
947512
answered 19 mins ago
LegionMammal978
947512
answered 19 mins ago
LegionMammal978
947512
947512
IMO this answer without number concatenation is better.
– Weather Vane
13 mins ago
@WeatherVane, but both are within the rules, so the sightly earlier answer gets the solution, but both get plus one.
– tom
5 mins ago
add a comment |Â
IMO this answer without number concatenation is better.
– Weather Vane
13 mins ago
@WeatherVane, but both are within the rules, so the sightly earlier answer gets the solution, but both get plus one.
– tom
5 mins ago
IMO this answer without number concatenation is better.
– Weather Vane
13 mins ago
IMO this answer without number concatenation is better.
– Weather Vane
13 mins ago
@WeatherVane, but both are within the rules, so the sightly earlier answer gets the solution, but both get plus one.
– tom
5 mins ago
@WeatherVane, but both are within the rules, so the sightly earlier answer gets the solution, but both get plus one.
– tom
5 mins ago
add a comment |Â
up vote
1
down vote
Can I have:
$56+sqrtsqrtdots3$
answered 21 mins ago
JonMark Perry
14.9k52972
That's a dotty answer. Interesting but is it in the rules?
– Weather Vane
14 mins ago
@WeatherVane it counts as an infinite number of operations, so gets plus one.
– tom
5 mins ago
@tom I see you changed the rules.
– Weather Vane
12 secs ago
add a comment |Â
up vote
1
down vote
Can I have:
$56+sqrtsqrtdots3$
answered 21 mins ago
JonMark Perry
14.9k52972
That's a dotty answer. Interesting but is it in the rules?
– Weather Vane
14 mins ago
@WeatherVane it counts as an infinite number of operations, so gets plus one.
– tom
5 mins ago
@tom I see you changed the rules.
– Weather Vane
12 secs ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Can I have:
$56+sqrtsqrtdots3$
answered 21 mins ago
JonMark Perry
14.9k52972
Can I have:
$56+sqrtsqrtdots3$
answered 21 mins ago
JonMark Perry
14.9k52972
answered 21 mins ago
JonMark Perry
14.9k52972
answered 21 mins ago
JonMark Perry
14.9k52972
answered 21 mins ago
JonMark Perry
14.9k52972
14.9k52972
That's a dotty answer. Interesting but is it in the rules?
– Weather Vane
14 mins ago
@WeatherVane it counts as an infinite number of operations, so gets plus one.
– tom
5 mins ago
@tom I see you changed the rules.
– Weather Vane
12 secs ago
add a comment |Â
That's a dotty answer. Interesting but is it in the rules?
– Weather Vane
14 mins ago
@WeatherVane it counts as an infinite number of operations, so gets plus one.
– tom
5 mins ago
@tom I see you changed the rules.
– Weather Vane
12 secs ago
That's a dotty answer. Interesting but is it in the rules?
– Weather Vane
14 mins ago
That's a dotty answer. Interesting but is it in the rules?
– Weather Vane
14 mins ago
@WeatherVane it counts as an infinite number of operations, so gets plus one.
– tom
5 mins ago
@WeatherVane it counts as an infinite number of operations, so gets plus one.
– tom
5 mins ago
@tom I see you changed the rules.
– Weather Vane
12 secs ago
@tom I see you changed the rules.
– Weather Vane
12 secs ago
add a comment |Â
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1
Tom, you're posting a lot of these puzzles and they're all basically the same and -- excuse my frankness! -- becoming rather boring. Maybe give them a rest for a while?
– Gareth McCaughan♦
25 mins ago
@GarethMcCaughan to be honest, some of them are very original, requires good logic deduction. but yes, would be good to give a rest for a while :)
– Oray
23 mins ago
@GarethMcCaughan you could say the same about the large number of Riley riddles. Some people like one kind of puzzle, but not the other, etc. Isn't the time to stop when no-one answers, or there are a lot of very quick correct answers?
– Weather Vane
19 mins ago
@GarethMcCaughan - point taken... I will retire from these riddles.
– tom
4 mins ago