I am caught when I was about to steal some candiesNumber swapping gameColored Pills True MADNESS -V2-16 Two Colored Line upFilling the board with 0s and 1sA pokemon puzzleCandy Button PaperGet 20 points from a target boardLots of children with Halloween candiesAnother variation of the game of Nim2019 gold coins to share
Do I have any obligations to my PhD supervisor's requests after I have graduated?
What is the highest voltage from the power supply a Raspberry Pi 3 B can handle without getting damaged?
Is "Busen" just the area between the breasts?
What constitutes a syllable?
How to mark the seams of UV maps to edit textures in external programs?
Mathematically modelling RC circuit with a linear input
Fantaisie Impromptu - Specific fingering issue
Why is it easier to balance a non-moving bike standing up than sitting down?
Very tricky nonogram - where to go next?
Can I enter the UK for 24 hours from a Schengen area, holding an Indian passport?
What is the oldest commercial MS-DOS program that can run on modern versions of Windows without third-party software?
Is there any proof that high saturation and contrast makes a picture more appealing in social media?
Drawing a second weapon as part of an attack?
What triggered jesuits' ban on infinitesimals in 1632?
Find All Possible Unique Combinations of Letters in a Word
Can the pre-order traversal of two different trees be the same even though they are different?
How does DC work with natural 20?
What does this Swiss black on yellow rectangular traffic sign with a symbol looking like a dart mean?
How do internally carried IR missiles acquire a lock?
Is the continuity test limit resistance of a multimeter standard?
Justifying Affordable Bespoke Spaceships
What is the meaning of "понаехать"?
What are the current battlegrounds for people’s “rights” in the UK?
Can you use one creature for both convoke and delve for Hogaak?
I am caught when I was about to steal some candies
Number swapping gameColored Pills True MADNESS -V2-16 Two Colored Line upFilling the board with 0s and 1sA pokemon puzzleCandy Button PaperGet 20 points from a target boardLots of children with Halloween candiesAnother variation of the game of Nim2019 gold coins to share
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I am a eight years old boy, I love candies, but my mom does not want me to eat any more candies because I start to have black teeth :( but I still love candies so...
Today when I was about to steal some candies from our candy box, my mom caught me! but strangely she wants to test my math skill and decided to give me a chance to get as many candies as I want if I win a game she just made up:
There are exactly 66 candies in this box, I will write all
numbers from 1 to 66 to your board (1,2,3,4...,66) you will choose a
number from the board and take that many candies from the box, but I
will wipe that number so you will not able to take that many candies
any more from the box. And also I will take candies from you to put back to the box with the
left numbers written on the board as you do and wipe that number. then it is your turn etc... We will play this game until
no number left on the board or no valid number of candies available to take from any of us (meaning if I have 20 candies, mom cannot try to take 21 candies from me)
If you can take the most candies from me at the end of the game, I will let you have that many candies just for today! otherwise, no candies and to your bed! Come on, tell me your first number!
Help me! I want to eat all candies,
is it possible? if not, how many can I get the most?
For the second case to end the game, I would like to clear with an example:
Let's say there are 50 candies in the candy box (mom's), and you have 16 candies: and somehow only 20,45,55 numbers are left on the board. if it is your mom's turn, she will not able to take any candies from you since the numbers on the board are bigger than the number of candies you have and the game ends there, have fun with your 16 candies :)
mathematics logical-deduction strategy
asked Jun 11 at 12:13
OrayOray
16.9k439171
$endgroup$
add a comment |
$begingroup$
I am a eight years old boy, I love candies, but my mom does not want me to eat any more candies because I start to have black teeth :( but I still love candies so...
Today when I was about to steal some candies from our candy box, my mom caught me! but strangely she wants to test my math skill and decided to give me a chance to get as many candies as I want if I win a game she just made up:
There are exactly 66 candies in this box, I will write all
numbers from 1 to 66 to your board (1,2,3,4...,66) you will choose a
number from the board and take that many candies from the box, but I
will wipe that number so you will not able to take that many candies
any more from the box. And also I will take candies from you to put back to the box with the
left numbers written on the board as you do and wipe that number. then it is your turn etc... We will play this game until
no number left on the board or no valid number of candies available to take from any of us (meaning if I have 20 candies, mom cannot try to take 21 candies from me)
If you can take the most candies from me at the end of the game, I will let you have that many candies just for today! otherwise, no candies and to your bed! Come on, tell me your first number!
Help me! I want to eat all candies,
is it possible? if not, how many can I get the most?
For the second case to end the game, I would like to clear with an example:
Let's say there are 50 candies in the candy box (mom's), and you have 16 candies: and somehow only 20,45,55 numbers are left on the board. if it is your mom's turn, she will not able to take any candies from you since the numbers on the board are bigger than the number of candies you have and the game ends there, have fun with your 16 candies :)
mathematics logical-deduction strategy
asked Jun 11 at 12:13
OrayOray
16.9k439171
$endgroup$
$begingroup$
Will mom take the candies from us and put them in the original box?
$endgroup$
– athin
Jun 11 at 12:20
1
$begingroup$
@athin mom takes that candies back to candy box
$endgroup$
– Oray
Jun 11 at 12:20
add a comment |
$begingroup$
I am a eight years old boy, I love candies, but my mom does not want me to eat any more candies because I start to have black teeth :( but I still love candies so...
Today when I was about to steal some candies from our candy box, my mom caught me! but strangely she wants to test my math skill and decided to give me a chance to get as many candies as I want if I win a game she just made up:
There are exactly 66 candies in this box, I will write all
numbers from 1 to 66 to your board (1,2,3,4...,66) you will choose a
number from the board and take that many candies from the box, but I
will wipe that number so you will not able to take that many candies
any more from the box. And also I will take candies from you to put back to the box with the
left numbers written on the board as you do and wipe that number. then it is your turn etc... We will play this game until
no number left on the board or no valid number of candies available to take from any of us (meaning if I have 20 candies, mom cannot try to take 21 candies from me)
If you can take the most candies from me at the end of the game, I will let you have that many candies just for today! otherwise, no candies and to your bed! Come on, tell me your first number!
Help me! I want to eat all candies,
is it possible? if not, how many can I get the most?
For the second case to end the game, I would like to clear with an example:
Let's say there are 50 candies in the candy box (mom's), and you have 16 candies: and somehow only 20,45,55 numbers are left on the board. if it is your mom's turn, she will not able to take any candies from you since the numbers on the board are bigger than the number of candies you have and the game ends there, have fun with your 16 candies :)
mathematics logical-deduction strategy
asked Jun 11 at 12:13
OrayOray
16.9k439171
$endgroup$
I am a eight years old boy, I love candies, but my mom does not want me to eat any more candies because I start to have black teeth :( but I still love candies so...
Today when I was about to steal some candies from our candy box, my mom caught me! but strangely she wants to test my math skill and decided to give me a chance to get as many candies as I want if I win a game she just made up:
There are exactly 66 candies in this box, I will write all
numbers from 1 to 66 to your board (1,2,3,4...,66) you will choose a
number from the board and take that many candies from the box, but I
will wipe that number so you will not able to take that many candies
any more from the box. And also I will take candies from you to put back to the box with the
left numbers written on the board as you do and wipe that number. then it is your turn etc... We will play this game until
no number left on the board or no valid number of candies available to take from any of us (meaning if I have 20 candies, mom cannot try to take 21 candies from me)
If you can take the most candies from me at the end of the game, I will let you have that many candies just for today! otherwise, no candies and to your bed! Come on, tell me your first number!
Help me! I want to eat all candies,
is it possible? if not, how many can I get the most?
For the second case to end the game, I would like to clear with an example:
Let's say there are 50 candies in the candy box (mom's), and you have 16 candies: and somehow only 20,45,55 numbers are left on the board. if it is your mom's turn, she will not able to take any candies from you since the numbers on the board are bigger than the number of candies you have and the game ends there, have fun with your 16 candies :)
mathematics logical-deduction strategy
mathematics logical-deduction strategy
asked Jun 11 at 12:13
OrayOray
16.9k439171
asked Jun 11 at 12:13
OrayOray
16.9k439171
edited Jun 11 at 12:21
Oray
asked Jun 11 at 12:13
OrayOray
16.9k439171
asked Jun 11 at 12:13
OrayOray
16.9k439171
asked Jun 11 at 12:13
OrayOray
16.9k439171
16.9k439171
$begingroup$
Will mom take the candies from us and put them in the original box?
$endgroup$
– athin
Jun 11 at 12:20
1
$begingroup$
@athin mom takes that candies back to candy box
$endgroup$
– Oray
Jun 11 at 12:20
add a comment |
$begingroup$
Will mom take the candies from us and put them in the original box?
$endgroup$
– athin
Jun 11 at 12:20
1
$begingroup$
@athin mom takes that candies back to candy box
$endgroup$
– Oray
Jun 11 at 12:20
$begingroup$
Will mom take the candies from us and put them in the original box?
$endgroup$
– athin
Jun 11 at 12:20
$begingroup$
Will mom take the candies from us and put them in the original box?
$endgroup$
– athin
Jun 11 at 12:20
1
1
$begingroup$
@athin mom takes that candies back to candy box
$endgroup$
– Oray
Jun 11 at 12:20
$begingroup$
@athin mom takes that candies back to candy box
$endgroup$
– Oray
Jun 11 at 12:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is one way to win almost all the candies:
Take $2$. Mom must now choose $1$, as that is the only available number less than or equal to $2$. You are left with $1$ candy.
Take $3$, giving you $4$ candies in total. Mom must now take $4$, as $1-3$ have already been played. You are left with none.
The above sequence eliminates the four lowest numbers from the board. You can however repeat the same procedure with the next four numbers (choose $6$, Mom is forced to choose $5$, choose $7$, Mom must choose $8$ leaving you with $6-5+7-8=0$).
Repeating this will eliminate all the numbers in sets of $4$, so can eliminate everything from $1$ to $64$, after which you still have no candies.
You final move is then to choose $65$. The only number on the board is $66$, which cannot be played and the game ends. You then end up with $65$ candies.
I think this solution is optimal, in the sense that you cannot end with more candies.
To get all $66$ candies, your last move must bring your total to $66$. However, as you did the first move and the number of numbers on the board is even, there must be at least one number on the board after your last move. If you had $66$ candies, all remaining numbers are valid moves for Mom, so she will be able to make a move and reduce your total. It is therefore impossible to end the game with all $66$ candies.
answered Jun 11 at 12:42
Jaap ScherphuisJaap Scherphuis
17.6k13177
$endgroup$
1
$begingroup$
Beat me by a minute - have the remaining candy as a prize :o)
$endgroup$
– GeeTee
Jun 11 at 12:45
1
$begingroup$
Sorry @GeeTee. Yum, candy.
$endgroup$
– Jaap Scherphuis
Jun 11 at 13:13
add a comment |
Your Answer
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%2f84935%2fi-am-caught-when-i-was-about-to-steal-some-candies%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is one way to win almost all the candies:
Take $2$. Mom must now choose $1$, as that is the only available number less than or equal to $2$. You are left with $1$ candy.
Take $3$, giving you $4$ candies in total. Mom must now take $4$, as $1-3$ have already been played. You are left with none.
The above sequence eliminates the four lowest numbers from the board. You can however repeat the same procedure with the next four numbers (choose $6$, Mom is forced to choose $5$, choose $7$, Mom must choose $8$ leaving you with $6-5+7-8=0$).
Repeating this will eliminate all the numbers in sets of $4$, so can eliminate everything from $1$ to $64$, after which you still have no candies.
You final move is then to choose $65$. The only number on the board is $66$, which cannot be played and the game ends. You then end up with $65$ candies.
I think this solution is optimal, in the sense that you cannot end with more candies.
To get all $66$ candies, your last move must bring your total to $66$. However, as you did the first move and the number of numbers on the board is even, there must be at least one number on the board after your last move. If you had $66$ candies, all remaining numbers are valid moves for Mom, so she will be able to make a move and reduce your total. It is therefore impossible to end the game with all $66$ candies.
answered Jun 11 at 12:42
Jaap ScherphuisJaap Scherphuis
17.6k13177
$endgroup$
1
$begingroup$
Beat me by a minute - have the remaining candy as a prize :o)
$endgroup$
– GeeTee
Jun 11 at 12:45
1
$begingroup$
Sorry @GeeTee. Yum, candy.
$endgroup$
– Jaap Scherphuis
Jun 11 at 13:13
add a comment |
$begingroup$
Here is one way to win almost all the candies:
Take $2$. Mom must now choose $1$, as that is the only available number less than or equal to $2$. You are left with $1$ candy.
Take $3$, giving you $4$ candies in total. Mom must now take $4$, as $1-3$ have already been played. You are left with none.
The above sequence eliminates the four lowest numbers from the board. You can however repeat the same procedure with the next four numbers (choose $6$, Mom is forced to choose $5$, choose $7$, Mom must choose $8$ leaving you with $6-5+7-8=0$).
Repeating this will eliminate all the numbers in sets of $4$, so can eliminate everything from $1$ to $64$, after which you still have no candies.
You final move is then to choose $65$. The only number on the board is $66$, which cannot be played and the game ends. You then end up with $65$ candies.
I think this solution is optimal, in the sense that you cannot end with more candies.
To get all $66$ candies, your last move must bring your total to $66$. However, as you did the first move and the number of numbers on the board is even, there must be at least one number on the board after your last move. If you had $66$ candies, all remaining numbers are valid moves for Mom, so she will be able to make a move and reduce your total. It is therefore impossible to end the game with all $66$ candies.
answered Jun 11 at 12:42
Jaap ScherphuisJaap Scherphuis
17.6k13177
$endgroup$
1
$begingroup$
Beat me by a minute - have the remaining candy as a prize :o)
$endgroup$
– GeeTee
Jun 11 at 12:45
1
$begingroup$
Sorry @GeeTee. Yum, candy.
$endgroup$
– Jaap Scherphuis
Jun 11 at 13:13
add a comment |
$begingroup$
Here is one way to win almost all the candies:
Take $2$. Mom must now choose $1$, as that is the only available number less than or equal to $2$. You are left with $1$ candy.
Take $3$, giving you $4$ candies in total. Mom must now take $4$, as $1-3$ have already been played. You are left with none.
The above sequence eliminates the four lowest numbers from the board. You can however repeat the same procedure with the next four numbers (choose $6$, Mom is forced to choose $5$, choose $7$, Mom must choose $8$ leaving you with $6-5+7-8=0$).
Repeating this will eliminate all the numbers in sets of $4$, so can eliminate everything from $1$ to $64$, after which you still have no candies.
You final move is then to choose $65$. The only number on the board is $66$, which cannot be played and the game ends. You then end up with $65$ candies.
I think this solution is optimal, in the sense that you cannot end with more candies.
To get all $66$ candies, your last move must bring your total to $66$. However, as you did the first move and the number of numbers on the board is even, there must be at least one number on the board after your last move. If you had $66$ candies, all remaining numbers are valid moves for Mom, so she will be able to make a move and reduce your total. It is therefore impossible to end the game with all $66$ candies.
answered Jun 11 at 12:42
Jaap ScherphuisJaap Scherphuis
17.6k13177
$endgroup$
Here is one way to win almost all the candies:
Take $2$. Mom must now choose $1$, as that is the only available number less than or equal to $2$. You are left with $1$ candy.
Take $3$, giving you $4$ candies in total. Mom must now take $4$, as $1-3$ have already been played. You are left with none.
The above sequence eliminates the four lowest numbers from the board. You can however repeat the same procedure with the next four numbers (choose $6$, Mom is forced to choose $5$, choose $7$, Mom must choose $8$ leaving you with $6-5+7-8=0$).
Repeating this will eliminate all the numbers in sets of $4$, so can eliminate everything from $1$ to $64$, after which you still have no candies.
You final move is then to choose $65$. The only number on the board is $66$, which cannot be played and the game ends. You then end up with $65$ candies.
I think this solution is optimal, in the sense that you cannot end with more candies.
To get all $66$ candies, your last move must bring your total to $66$. However, as you did the first move and the number of numbers on the board is even, there must be at least one number on the board after your last move. If you had $66$ candies, all remaining numbers are valid moves for Mom, so she will be able to make a move and reduce your total. It is therefore impossible to end the game with all $66$ candies.
answered Jun 11 at 12:42
Jaap ScherphuisJaap Scherphuis
17.6k13177
edited Jun 11 at 13:12
answered Jun 11 at 12:42
Jaap ScherphuisJaap Scherphuis
17.6k13177
answered Jun 11 at 12:42
Jaap ScherphuisJaap Scherphuis
17.6k13177
answered Jun 11 at 12:42
Jaap ScherphuisJaap Scherphuis
17.6k13177
17.6k13177
1
$begingroup$
Beat me by a minute - have the remaining candy as a prize :o)
$endgroup$
– GeeTee
Jun 11 at 12:45
1
$begingroup$
Sorry @GeeTee. Yum, candy.
$endgroup$
– Jaap Scherphuis
Jun 11 at 13:13
add a comment |
1
$begingroup$
Beat me by a minute - have the remaining candy as a prize :o)
$endgroup$
– GeeTee
Jun 11 at 12:45
1
$begingroup$
Sorry @GeeTee. Yum, candy.
$endgroup$
– Jaap Scherphuis
Jun 11 at 13:13
1
1
$begingroup$
Beat me by a minute - have the remaining candy as a prize :o)
$endgroup$
– GeeTee
Jun 11 at 12:45
$begingroup$
Beat me by a minute - have the remaining candy as a prize :o)
$endgroup$
– GeeTee
Jun 11 at 12:45
1
1
$begingroup$
Sorry @GeeTee. Yum, candy.
$endgroup$
– Jaap Scherphuis
Jun 11 at 13:13
$begingroup$
Sorry @GeeTee. Yum, candy.
$endgroup$
– Jaap Scherphuis
Jun 11 at 13:13
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%2f84935%2fi-am-caught-when-i-was-about-to-steal-some-candies%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
$begingroup$
Will mom take the candies from us and put them in the original box?
$endgroup$
– athin
Jun 11 at 12:20
1
$begingroup$
@athin mom takes that candies back to candy box
$endgroup$
– Oray
Jun 11 at 12:20