Number of seconds in 6 weeks1 2 3 4 5 6 7 8 9 = 100The game of SevensExpress the number $2015$ using only the digit $2$ twiceFill in the operators to make $7 circ 8 circ 7 circ 7 circ 8 circ 3 = 100$10 9 8 7 6 5 4 3 2 1 = 2016Use a circuit to multiply two resistancesMake 11 from five identical digitsA (Simple) Arithmetic puzzle10 9 8 7 6 5 4 3 2 1 = 2017Construct $sqrt3$ using every natural number
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Number of seconds in 6 weeks
1 2 3 4 5 6 7 8 9 = 100The game of SevensExpress the number $2015$ using only the digit $2$ twiceFill in the operators to make $7 circ 8 circ 7 circ 7 circ 8 circ 3 = 100$10 9 8 7 6 5 4 3 2 1 = 2016Use a circuit to multiply two resistancesMake 11 from five identical digitsA (Simple) Arithmetic puzzle10 9 8 7 6 5 4 3 2 1 = 2017Construct $sqrt3$ using every natural number
$begingroup$
Using only 3 characters (any single digit or one of the following operators: $ +,, -,, times,, div,, ! $), construct an expression that evaluates to the number of seconds in 6 weeks.
mathematics
edited 2 hours ago
PiIsNot3
3,140643
asked 3 hours ago
UvcUvc
212
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
Using only 3 characters (any single digit or one of the following operators: $ +,, -,, times,, div,, ! $), construct an expression that evaluates to the number of seconds in 6 weeks.
mathematics
edited 2 hours ago
PiIsNot3
3,140643
asked 3 hours ago
UvcUvc
212
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
8
$begingroup$
Are you sure it's not 6 weeks?
$endgroup$
– noedne
3 hours ago
2
$begingroup$
I second noedne's question.
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Sorry about that..it should be 6 weeks
$endgroup$
– Uvc
2 hours ago
1
$begingroup$
^I added this information in to the original post. In the future, if you post questions like these, please make sure that all possible specifications are included. Thanks! :)
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
$endgroup$
– Uvc
2 hours ago
|
show 1 more comment
$begingroup$
Using only 3 characters (any single digit or one of the following operators: $ +,, -,, times,, div,, ! $), construct an expression that evaluates to the number of seconds in 6 weeks.
mathematics
edited 2 hours ago
PiIsNot3
3,140643
asked 3 hours ago
UvcUvc
212
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Using only 3 characters (any single digit or one of the following operators: $ +,, -,, times,, div,, ! $), construct an expression that evaluates to the number of seconds in 6 weeks.
mathematics
mathematics
edited 2 hours ago
PiIsNot3
3,140643
asked 3 hours ago
UvcUvc
212
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
PiIsNot3
3,140643
asked 3 hours ago
UvcUvc
212
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
PiIsNot3
3,140643
edited 2 hours ago
PiIsNot3
3,140643
edited 2 hours ago
PiIsNot3
3,140643
3,140643
asked 3 hours ago
UvcUvc
212
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
UvcUvc
212
asked 3 hours ago
UvcUvc
212
212
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
8
$begingroup$
Are you sure it's not 6 weeks?
$endgroup$
– noedne
3 hours ago
2
$begingroup$
I second noedne's question.
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Sorry about that..it should be 6 weeks
$endgroup$
– Uvc
2 hours ago
1
$begingroup$
^I added this information in to the original post. In the future, if you post questions like these, please make sure that all possible specifications are included. Thanks! :)
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
$endgroup$
– Uvc
2 hours ago
|
show 1 more comment
8
$begingroup$
Are you sure it's not 6 weeks?
$endgroup$
– noedne
3 hours ago
2
$begingroup$
I second noedne's question.
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Sorry about that..it should be 6 weeks
$endgroup$
– Uvc
2 hours ago
1
$begingroup$
^I added this information in to the original post. In the future, if you post questions like these, please make sure that all possible specifications are included. Thanks! :)
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
$endgroup$
– Uvc
2 hours ago
8
8
$begingroup$
Are you sure it's not 6 weeks?
$endgroup$
– noedne
3 hours ago
$begingroup$
Are you sure it's not 6 weeks?
$endgroup$
– noedne
3 hours ago
2
2
$begingroup$
I second noedne's question.
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
I second noedne's question.
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Sorry about that..it should be 6 weeks
$endgroup$
– Uvc
2 hours ago
$begingroup$
Sorry about that..it should be 6 weeks
$endgroup$
– Uvc
2 hours ago
1
1
$begingroup$
^I added this information in to the original post. In the future, if you post questions like these, please make sure that all possible specifications are included. Thanks! :)
$endgroup$
– PiIsNot3
2 hours ago
$begingroup$
^I added this information in to the original post. In the future, if you post questions like these, please make sure that all possible specifications are included. Thanks! :)
$endgroup$
– PiIsNot3
2 hours ago
1
1
$begingroup$
Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
$endgroup$
– Uvc
2 hours ago
$begingroup$
Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
$endgroup$
– Uvc
2 hours ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
One answer is:
'$10!$' = 3628800 = (60*60*24*7*6)
And for fun, the original question accidentally asked to solve for 3 weeks. An argument could be made that the following works:
'$9!5$' evaluates to (9! * 5) = 1814400 = (60*60*24*7*3)
answered 2 hours ago
TwoBitOperationTwoBitOperation
8,73411667
$endgroup$
$begingroup$
I'm not entirely sure about your solution for 3 weeks, since it implicitly assumes that two numbers next to each other is multiplication, which is not always the case (concatenation is another possibility). But kudos for trying to find a way to do it!
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
@PiIsNot3 When an operator is involved, concatenation is hardly an option. The real problem with this solution is that it could be $9$ times the subfactorial of $5$...
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier sure, that's another possibility, but the main point still stands that putting those values next to each other is ambiguous (which is probably why the question got changed from 3 weeks to 6 weeks)
$endgroup$
– PiIsNot3
1 hour ago
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One answer is:
'$10!$' = 3628800 = (60*60*24*7*6)
And for fun, the original question accidentally asked to solve for 3 weeks. An argument could be made that the following works:
'$9!5$' evaluates to (9! * 5) = 1814400 = (60*60*24*7*3)
answered 2 hours ago
TwoBitOperationTwoBitOperation
8,73411667
$endgroup$
$begingroup$
I'm not entirely sure about your solution for 3 weeks, since it implicitly assumes that two numbers next to each other is multiplication, which is not always the case (concatenation is another possibility). But kudos for trying to find a way to do it!
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
@PiIsNot3 When an operator is involved, concatenation is hardly an option. The real problem with this solution is that it could be $9$ times the subfactorial of $5$...
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier sure, that's another possibility, but the main point still stands that putting those values next to each other is ambiguous (which is probably why the question got changed from 3 weeks to 6 weeks)
$endgroup$
– PiIsNot3
1 hour ago
add a comment |
$begingroup$
One answer is:
'$10!$' = 3628800 = (60*60*24*7*6)
And for fun, the original question accidentally asked to solve for 3 weeks. An argument could be made that the following works:
'$9!5$' evaluates to (9! * 5) = 1814400 = (60*60*24*7*3)
answered 2 hours ago
TwoBitOperationTwoBitOperation
8,73411667
$endgroup$
$begingroup$
I'm not entirely sure about your solution for 3 weeks, since it implicitly assumes that two numbers next to each other is multiplication, which is not always the case (concatenation is another possibility). But kudos for trying to find a way to do it!
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
@PiIsNot3 When an operator is involved, concatenation is hardly an option. The real problem with this solution is that it could be $9$ times the subfactorial of $5$...
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier sure, that's another possibility, but the main point still stands that putting those values next to each other is ambiguous (which is probably why the question got changed from 3 weeks to 6 weeks)
$endgroup$
– PiIsNot3
1 hour ago
add a comment |
$begingroup$
One answer is:
'$10!$' = 3628800 = (60*60*24*7*6)
And for fun, the original question accidentally asked to solve for 3 weeks. An argument could be made that the following works:
'$9!5$' evaluates to (9! * 5) = 1814400 = (60*60*24*7*3)
answered 2 hours ago
TwoBitOperationTwoBitOperation
8,73411667
$endgroup$
One answer is:
'$10!$' = 3628800 = (60*60*24*7*6)
And for fun, the original question accidentally asked to solve for 3 weeks. An argument could be made that the following works:
'$9!5$' evaluates to (9! * 5) = 1814400 = (60*60*24*7*3)
answered 2 hours ago
TwoBitOperationTwoBitOperation
8,73411667
edited 2 hours ago
answered 2 hours ago
TwoBitOperationTwoBitOperation
8,73411667
answered 2 hours ago
TwoBitOperationTwoBitOperation
8,73411667
answered 2 hours ago
TwoBitOperationTwoBitOperation
8,73411667
8,73411667
$begingroup$
I'm not entirely sure about your solution for 3 weeks, since it implicitly assumes that two numbers next to each other is multiplication, which is not always the case (concatenation is another possibility). But kudos for trying to find a way to do it!
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
@PiIsNot3 When an operator is involved, concatenation is hardly an option. The real problem with this solution is that it could be $9$ times the subfactorial of $5$...
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier sure, that's another possibility, but the main point still stands that putting those values next to each other is ambiguous (which is probably why the question got changed from 3 weeks to 6 weeks)
$endgroup$
– PiIsNot3
1 hour ago
add a comment |
$begingroup$
I'm not entirely sure about your solution for 3 weeks, since it implicitly assumes that two numbers next to each other is multiplication, which is not always the case (concatenation is another possibility). But kudos for trying to find a way to do it!
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
@PiIsNot3 When an operator is involved, concatenation is hardly an option. The real problem with this solution is that it could be $9$ times the subfactorial of $5$...
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier sure, that's another possibility, but the main point still stands that putting those values next to each other is ambiguous (which is probably why the question got changed from 3 weeks to 6 weeks)
$endgroup$
– PiIsNot3
1 hour ago
$begingroup$
I'm not entirely sure about your solution for 3 weeks, since it implicitly assumes that two numbers next to each other is multiplication, which is not always the case (concatenation is another possibility). But kudos for trying to find a way to do it!
$endgroup$
– PiIsNot3
2 hours ago
$begingroup$
I'm not entirely sure about your solution for 3 weeks, since it implicitly assumes that two numbers next to each other is multiplication, which is not always the case (concatenation is another possibility). But kudos for trying to find a way to do it!
$endgroup$
– PiIsNot3
2 hours ago
1
1
$begingroup$
@PiIsNot3 When an operator is involved, concatenation is hardly an option. The real problem with this solution is that it could be $9$ times the subfactorial of $5$...
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@PiIsNot3 When an operator is involved, concatenation is hardly an option. The real problem with this solution is that it could be $9$ times the subfactorial of $5$...
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier sure, that's another possibility, but the main point still stands that putting those values next to each other is ambiguous (which is probably why the question got changed from 3 weeks to 6 weeks)
$endgroup$
– PiIsNot3
1 hour ago
$begingroup$
@ArnaudMortier sure, that's another possibility, but the main point still stands that putting those values next to each other is ambiguous (which is probably why the question got changed from 3 weeks to 6 weeks)
$endgroup$
– PiIsNot3
1 hour ago
add a comment |
Uvc is a new contributor. Be nice, and check out our Code of Conduct.
Uvc is a new contributor. Be nice, and check out our Code of Conduct.
Uvc is a new contributor. Be nice, and check out our Code of Conduct.
Uvc is a new contributor. Be nice, and check out our Code of Conduct.
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8
$begingroup$
Are you sure it's not 6 weeks?
$endgroup$
– noedne
3 hours ago
2
$begingroup$
I second noedne's question.
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Sorry about that..it should be 6 weeks
$endgroup$
– Uvc
2 hours ago
1
$begingroup$
^I added this information in to the original post. In the future, if you post questions like these, please make sure that all possible specifications are included. Thanks! :)
$endgroup$
– PiIsNot3
2 hours ago
1
$begingroup$
Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
$endgroup$
– Uvc
2 hours ago