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













2












$begingroup$


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.










share|improve this question









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.







$endgroup$











  • $begingroup$
    Usual..+,-,/,!,x
    $endgroup$
    – Uvc
    Apr 30 at 18:55






  • 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
    Apr 30 at 19:02







  • 1




    $begingroup$
    Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
    $endgroup$
    – Uvc
    Apr 30 at 19:05










  • $begingroup$
    is × multiplication?
    $endgroup$
    – Mom344
    May 1 at 0:35















2












$begingroup$


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.










share|improve this question









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.







$endgroup$











  • $begingroup$
    Usual..+,-,/,!,x
    $endgroup$
    – Uvc
    Apr 30 at 18:55






  • 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
    Apr 30 at 19:02







  • 1




    $begingroup$
    Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
    $endgroup$
    – Uvc
    Apr 30 at 19:05










  • $begingroup$
    is × multiplication?
    $endgroup$
    – Mom344
    May 1 at 0:35













2












2








2





$begingroup$


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.










share|improve this question









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






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited Apr 30 at 19:01









PiIsNot3

3,572847




3,572847






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 Apr 30 at 18:28









UvcUvc

593




593




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.











  • $begingroup$
    Usual..+,-,/,!,x
    $endgroup$
    – Uvc
    Apr 30 at 18:55






  • 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
    Apr 30 at 19:02







  • 1




    $begingroup$
    Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
    $endgroup$
    – Uvc
    Apr 30 at 19:05










  • $begingroup$
    is × multiplication?
    $endgroup$
    – Mom344
    May 1 at 0:35
















  • $begingroup$
    Usual..+,-,/,!,x
    $endgroup$
    – Uvc
    Apr 30 at 18:55






  • 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
    Apr 30 at 19:02







  • 1




    $begingroup$
    Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
    $endgroup$
    – Uvc
    Apr 30 at 19:05










  • $begingroup$
    is × multiplication?
    $endgroup$
    – Mom344
    May 1 at 0:35















$begingroup$
Usual..+,-,/,!,x
$endgroup$
– Uvc
Apr 30 at 18:55




$begingroup$
Usual..+,-,/,!,x
$endgroup$
– Uvc
Apr 30 at 18:55




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
Apr 30 at 19:02





$begingroup$
^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
Apr 30 at 19:02





1




1




$begingroup$
Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
$endgroup$
– Uvc
Apr 30 at 19:05




$begingroup$
Thanks for clarifying the question..I am a novice to the site and hopefully improve as I go along.
$endgroup$
– Uvc
Apr 30 at 19:05












$begingroup$
is × multiplication?
$endgroup$
– Mom344
May 1 at 0:35




$begingroup$
is × multiplication?
$endgroup$
– Mom344
May 1 at 0:35










1 Answer
1






active

oldest

votes


















9












$begingroup$

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)







share|improve this answer











$endgroup$












  • $begingroup$
    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
    Apr 30 at 19:14






  • 2




    $begingroup$
    @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
    Apr 30 at 20:01











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









9












$begingroup$

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)







share|improve this answer











$endgroup$












  • $begingroup$
    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
    Apr 30 at 19:14






  • 2




    $begingroup$
    @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
    Apr 30 at 20:01















9












$begingroup$

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)







share|improve this answer











$endgroup$












  • $begingroup$
    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
    Apr 30 at 19:14






  • 2




    $begingroup$
    @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
    Apr 30 at 20:01













9












9








9





$begingroup$

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)







share|improve this answer











$endgroup$



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)








share|improve this answer














share|improve this answer



share|improve this answer








edited Apr 30 at 19:11

























answered Apr 30 at 18:45









TwoBitOperationTwoBitOperation

8,83411667




8,83411667











  • $begingroup$
    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
    Apr 30 at 19:14






  • 2




    $begingroup$
    @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
    Apr 30 at 20:01
















  • $begingroup$
    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
    Apr 30 at 19:14






  • 2




    $begingroup$
    @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
    Apr 30 at 20:01















$begingroup$
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
Apr 30 at 19:14




$begingroup$
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
Apr 30 at 19:14




2




2




$begingroup$
@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
Apr 30 at 20:01




$begingroup$
@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
Apr 30 at 20:01










Uvc is a new contributor. Be nice, and check out our Code of Conduct.









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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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