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
New contributor
$endgroup$
add a 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
New contributor
$endgroup$
$begingroup$
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
add a 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
New contributor
$endgroup$
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
New contributor
New contributor
edited Apr 30 at 19:01
PiIsNot3
3,572847
3,572847
New contributor
asked Apr 30 at 18:28
UvcUvc
593
593
New contributor
New contributor
$begingroup$
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
add a comment |
$begingroup$
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
add a 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)
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
1
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oldest
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active
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active
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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)
$endgroup$
$begingroup$
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
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)
$endgroup$
$begingroup$
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
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)
$endgroup$
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)
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
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
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
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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$begingroup$
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