1 0 1 0 1 0 1 0 1 0 1
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 $requireencloseenclosehorizontalstrike10101 $ (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 $requireencloseenclosehorizontalstrike10101 $ (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♦
Mar 3 at 0:01
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
Mar 3 at 0:04
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
Mar 3 at 0:14
$begingroup$
That's an interesting title you got there.
$endgroup$
– Rand al'Thor
Mar 3 at 8:45
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 $requireencloseenclosehorizontalstrike10101 $ (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 $requireencloseenclosehorizontalstrike10101 $ (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 Mar 3 at 0:25
Rubio
asked Mar 2 at 23:50
Rubio♦Rubio
30.4k567188
30.4k567188
$begingroup$
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♦
Mar 3 at 0:01
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
Mar 3 at 0:04
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
Mar 3 at 0:14
$begingroup$
That's an interesting title you got there.
$endgroup$
– Rand al'Thor
Mar 3 at 8:45
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♦
Mar 3 at 0:01
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
Mar 3 at 0:04
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
Mar 3 at 0:14
$begingroup$
That's an interesting title you got there.
$endgroup$
– Rand al'Thor
Mar 3 at 8:45
$begingroup$
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♦
Mar 3 at 0:01
$begingroup$
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♦
Mar 3 at 0:01
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
Mar 3 at 0:04
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
Mar 3 at 0:04
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
Mar 3 at 0:14
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
Mar 3 at 0:14
$begingroup$
That's an interesting title you got there.
$endgroup$
– Rand al'Thor
Mar 3 at 8:45
$begingroup$
That's an interesting title you got there.
$endgroup$
– Rand al'Thor
Mar 3 at 8:45
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
Mar 3 at 0:07
$begingroup$
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♦
Mar 3 at 0:09
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
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "559"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f80169%2f1-0-1-0-1-0-1-0-1-0-1%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Mar 3 at 0:07
$begingroup$
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♦
Mar 3 at 0:09
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
Mar 3 at 0:07
$begingroup$
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♦
Mar 3 at 0:09
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 Mar 3 at 0:12
answered Mar 3 at 0:03
Gareth McCaughan♦Gareth McCaughan
65.5k3166256
65.5k3166256
$begingroup$
"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
Mar 3 at 0:07
$begingroup$
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♦
Mar 3 at 0:09
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
Mar 3 at 0:07
$begingroup$
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♦
Mar 3 at 0:09
$begingroup$
"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
Mar 3 at 0:07
$begingroup$
"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
Mar 3 at 0:07
$begingroup$
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♦
Mar 3 at 0:09
$begingroup$
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♦
Mar 3 at 0:09
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 Mar 3 at 0:42
EKonsEKons
1,045826
1,045826
add a comment |
add a comment |
Thanks for contributing an answer to Puzzling Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f80169%2f1-0-1-0-1-0-1-0-1-0-1%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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♦
Mar 3 at 0:01
$begingroup$
Further clarification: (3) the number 0 is allowed despite the "no leading zeroes" rule?
$endgroup$
– Gareth McCaughan♦
Mar 3 at 0:04
$begingroup$
Yes to all the above.
$endgroup$
– Rubio♦
Mar 3 at 0:14
$begingroup$
That's an interesting title you got there.
$endgroup$
– Rand al'Thor
Mar 3 at 8:45