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1 0 1 0 1 0 1 0 1 0 1
Making π from 1 2 3 4 5 6 7 8 9Number formation with digits givenA truly amazing way of making the number 2016A truly amazing way of making every possible positive integerRiddle with functionsExpressing numbers using 0, 1, 2, 3, and 4All digits with one operationMake numbers 1-100 with only four 8sMake 38, 44, 46 using 2,3,8,7?Create all numbers from 1-100 by using 1,3,3,7
$begingroup$
Start with the digits:
$1 0 1 0 1 0 1 0 1 0 1$
You may then add (only!) these simple arithmetic operators:
$+ - times div$
You may also remove the space between no more than 3 adjacent digits, to make larger numbers:
$1 0 1$ can be concatenated to $101$, for example
$1 0 1 0 1$ cannot be concatenated to $require{enclose}enclose{horizontalstrike}{10101} $ (too many digits)
You must then insert one equal sign $ = $ between two numbers to form a valid and correct equation.
What is the fewest operators you can add to form a valid equation?
[Added bonus question]
What is the fewest operators you can add to form a valid equation if operators only appear on one side of the equal sign?
Unary negation $-1$ is allowed.
Unary plus $+1$ is not allowed.
Binary addition, subtraction, multiplication, and division are allowed.
No other operators are allowed.
No decimal points, and no rounding.
Leading zeroes on numbers are not allowed.
Base 10 only.
Order of operations:
What Perl uses. Basically, BO(DM)(AS):
• left to right
• multiplication and division done together in one pass
• addition and subtraction done together in a lower priority pass
You may add parentheses if required to reorder evaluation. Each pair counts as an operator.
no-computers formation-of-numbers arithmetic
$endgroup$
add a comment |
$begingroup$
Start with the digits:
$1 0 1 0 1 0 1 0 1 0 1$
You may then add (only!) these simple arithmetic operators:
$+ - times div$
You may also remove the space between no more than 3 adjacent digits, to make larger numbers:
$1 0 1$ can be concatenated to $101$, for example
$1 0 1 0 1$ cannot be concatenated to $require{enclose}enclose{horizontalstrike}{10101} $ (too many digits)
You must then insert one equal sign $ = $ between two numbers to form a valid and correct equation.
What is the fewest operators you can add to form a valid equation?
[Added bonus question]
What is the fewest operators you can add to form a valid equation if operators only appear on one side of the equal sign?
Unary negation $-1$ is allowed.
Unary plus $+1$ is not allowed.
Binary addition, subtraction, multiplication, and division are allowed.
No other operators are allowed.
No decimal points, and no rounding.
Leading zeroes on numbers are not allowed.
Base 10 only.
Order of operations:
What Perl uses. Basically, BO(DM)(AS):
• left to right
• multiplication and division done together in one pass
• addition and subtraction done together in a lower priority pass
You may add parentheses if required to reorder evaluation. Each pair counts as an operator.
no-computers formation-of-numbers arithmetic
$endgroup$
$begingroup$
Just to be clear: (1) you're allowed any number of arithmetic operators, not just one of each?, (2) you can do any number of space-removals, just not so as to make any single number longer than 3 digits?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
2 hours ago
add a comment |
$begingroup$
Start with the digits:
$1 0 1 0 1 0 1 0 1 0 1$
You may then add (only!) these simple arithmetic operators:
$+ - times div$
You may also remove the space between no more than 3 adjacent digits, to make larger numbers:
$1 0 1$ can be concatenated to $101$, for example
$1 0 1 0 1$ cannot be concatenated to $require{enclose}enclose{horizontalstrike}{10101} $ (too many digits)
You must then insert one equal sign $ = $ between two numbers to form a valid and correct equation.
What is the fewest operators you can add to form a valid equation?
[Added bonus question]
What is the fewest operators you can add to form a valid equation if operators only appear on one side of the equal sign?
Unary negation $-1$ is allowed.
Unary plus $+1$ is not allowed.
Binary addition, subtraction, multiplication, and division are allowed.
No other operators are allowed.
No decimal points, and no rounding.
Leading zeroes on numbers are not allowed.
Base 10 only.
Order of operations:
What Perl uses. Basically, BO(DM)(AS):
• left to right
• multiplication and division done together in one pass
• addition and subtraction done together in a lower priority pass
You may add parentheses if required to reorder evaluation. Each pair counts as an operator.
no-computers formation-of-numbers arithmetic
$endgroup$
Start with the digits:
$1 0 1 0 1 0 1 0 1 0 1$
You may then add (only!) these simple arithmetic operators:
$+ - times div$
You may also remove the space between no more than 3 adjacent digits, to make larger numbers:
$1 0 1$ can be concatenated to $101$, for example
$1 0 1 0 1$ cannot be concatenated to $require{enclose}enclose{horizontalstrike}{10101} $ (too many digits)
You must then insert one equal sign $ = $ between two numbers to form a valid and correct equation.
What is the fewest operators you can add to form a valid equation?
[Added bonus question]
What is the fewest operators you can add to form a valid equation if operators only appear on one side of the equal sign?
Unary negation $-1$ is allowed.
Unary plus $+1$ is not allowed.
Binary addition, subtraction, multiplication, and division are allowed.
No other operators are allowed.
No decimal points, and no rounding.
Leading zeroes on numbers are not allowed.
Base 10 only.
Order of operations:
What Perl uses. Basically, BO(DM)(AS):
• left to right
• multiplication and division done together in one pass
• addition and subtraction done together in a lower priority pass
You may add parentheses if required to reorder evaluation. Each pair counts as an operator.
no-computers formation-of-numbers arithmetic
no-computers formation-of-numbers arithmetic
edited 2 hours ago
Rubio
asked 2 hours ago
Rubio♦Rubio
29k565178
29k565178
$begingroup$
Just to be clear: (1) you're allowed any number of arithmetic operators, not just one of each?, (2) you can do any number of space-removals, just not so as to make any single number longer than 3 digits?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
2 hours ago
add a comment |
$begingroup$
Just to be clear: (1) you're allowed any number of arithmetic operators, not just one of each?, (2) you can do any number of space-removals, just not so as to make any single number longer than 3 digits?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
2 hours ago
$begingroup$
Just to be clear: (1) you're allowed any number of arithmetic operators, not just one of each?, (2) you can do any number of space-removals, just not so as to make any single number longer than 3 digits?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Just to be clear: (1) you're allowed any number of arithmetic operators, not just one of each?, (2) you can do any number of space-removals, just not so as to make any single number longer than 3 digits?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
2 hours ago
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is a simple, but optimal, solution.
10+101+0=10+101
uses a total of
three additions, one =, and no other operators.
The best one could hope for is
two operators (and one =) -- since no number can be longer than 3 digits, we have to break up the string of digits in at least three places. So the above can't be improved much.
And in fact,
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
$endgroup$
$begingroup$
"The best one could hope for is" exactly what I'm thinking about right now. ;-) The solution you've provided (and its trivial variants) is the easy one.
$endgroup$
– EKons
2 hours ago
$begingroup$
I actually think it's impossible to do better than the easy one, but I haven't checked my reasoning yet or even written it down explicitly... [EDITED to add:] Yup, seems to be correct.
$endgroup$
– Gareth McCaughan♦
2 hours ago
add a comment |
$begingroup$
To answer the bonus question, you'll need:
$10-10+101+0=101$
This is optimal for the reason given in Gareth's answer by himself:
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
Of course, the main solution has already been given.
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a simple, but optimal, solution.
10+101+0=10+101
uses a total of
three additions, one =, and no other operators.
The best one could hope for is
two operators (and one =) -- since no number can be longer than 3 digits, we have to break up the string of digits in at least three places. So the above can't be improved much.
And in fact,
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
$endgroup$
$begingroup$
"The best one could hope for is" exactly what I'm thinking about right now. ;-) The solution you've provided (and its trivial variants) is the easy one.
$endgroup$
– EKons
2 hours ago
$begingroup$
I actually think it's impossible to do better than the easy one, but I haven't checked my reasoning yet or even written it down explicitly... [EDITED to add:] Yup, seems to be correct.
$endgroup$
– Gareth McCaughan♦
2 hours ago
add a comment |
$begingroup$
Here is a simple, but optimal, solution.
10+101+0=10+101
uses a total of
three additions, one =, and no other operators.
The best one could hope for is
two operators (and one =) -- since no number can be longer than 3 digits, we have to break up the string of digits in at least three places. So the above can't be improved much.
And in fact,
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
$endgroup$
$begingroup$
"The best one could hope for is" exactly what I'm thinking about right now. ;-) The solution you've provided (and its trivial variants) is the easy one.
$endgroup$
– EKons
2 hours ago
$begingroup$
I actually think it's impossible to do better than the easy one, but I haven't checked my reasoning yet or even written it down explicitly... [EDITED to add:] Yup, seems to be correct.
$endgroup$
– Gareth McCaughan♦
2 hours ago
add a comment |
$begingroup$
Here is a simple, but optimal, solution.
10+101+0=10+101
uses a total of
three additions, one =, and no other operators.
The best one could hope for is
two operators (and one =) -- since no number can be longer than 3 digits, we have to break up the string of digits in at least three places. So the above can't be improved much.
And in fact,
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
$endgroup$
Here is a simple, but optimal, solution.
10+101+0=10+101
uses a total of
three additions, one =, and no other operators.
The best one could hope for is
two operators (and one =) -- since no number can be longer than 3 digits, we have to break up the string of digits in at least three places. So the above can't be improved much.
And in fact,
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
edited 2 hours ago
answered 2 hours ago
Gareth McCaughan♦Gareth McCaughan
63.8k3163249
63.8k3163249
$begingroup$
"The best one could hope for is" exactly what I'm thinking about right now. ;-) The solution you've provided (and its trivial variants) is the easy one.
$endgroup$
– EKons
2 hours ago
$begingroup$
I actually think it's impossible to do better than the easy one, but I haven't checked my reasoning yet or even written it down explicitly... [EDITED to add:] Yup, seems to be correct.
$endgroup$
– Gareth McCaughan♦
2 hours ago
add a comment |
$begingroup$
"The best one could hope for is" exactly what I'm thinking about right now. ;-) The solution you've provided (and its trivial variants) is the easy one.
$endgroup$
– EKons
2 hours ago
$begingroup$
I actually think it's impossible to do better than the easy one, but I haven't checked my reasoning yet or even written it down explicitly... [EDITED to add:] Yup, seems to be correct.
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
"The best one could hope for is" exactly what I'm thinking about right now. ;-) The solution you've provided (and its trivial variants) is the easy one.
$endgroup$
– EKons
2 hours ago
$begingroup$
"The best one could hope for is" exactly what I'm thinking about right now. ;-) The solution you've provided (and its trivial variants) is the easy one.
$endgroup$
– EKons
2 hours ago
$begingroup$
I actually think it's impossible to do better than the easy one, but I haven't checked my reasoning yet or even written it down explicitly... [EDITED to add:] Yup, seems to be correct.
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
I actually think it's impossible to do better than the easy one, but I haven't checked my reasoning yet or even written it down explicitly... [EDITED to add:] Yup, seems to be correct.
$endgroup$
– Gareth McCaughan♦
2 hours ago
add a comment |
$begingroup$
To answer the bonus question, you'll need:
$10-10+101+0=101$
This is optimal for the reason given in Gareth's answer by himself:
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
Of course, the main solution has already been given.
$endgroup$
add a comment |
$begingroup$
To answer the bonus question, you'll need:
$10-10+101+0=101$
This is optimal for the reason given in Gareth's answer by himself:
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
Of course, the main solution has already been given.
$endgroup$
add a comment |
$begingroup$
To answer the bonus question, you'll need:
$10-10+101+0=101$
This is optimal for the reason given in Gareth's answer by himself:
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
Of course, the main solution has already been given.
$endgroup$
To answer the bonus question, you'll need:
$10-10+101+0=101$
This is optimal for the reason given in Gareth's answer by himself:
we have to break in at least four places, so the above is optimal. Why? Because if we ever break after a 1 then the next number starts with a 0 and hence must be exactly a 0; if there is a single-digit number then the remaining 10 digits must still form >3 groups (because no group is longer than 3), so we get at least 5 groups in all, hence at least 4 group-breakers, hence at least 3 operators. On the other hand, if we never break after a 1 then all our groups other than perhaps the last one are of even size, hence of size at most 2, which again means at least 5 groups in all.
Of course, the main solution has already been given.
answered 1 hour ago
EKonsEKons
1,015826
1,015826
add a comment |
add a comment |
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$begingroup$
Just to be clear: (1) you're allowed any number of arithmetic operators, not just one of each?, (2) you can do any number of space-removals, just not so as to make any single number longer than 3 digits?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
2 hours ago
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
2 hours ago