Picking marbles
We have 15 urns each of them having a different number of marbles, from 1 to 15. We start by picking the same number of marbles from each of the urns we choose. We repeat the process until we have picked all marbles. What is the minimum number of days we can finish picking all marbles? Just to clarify that it is not necessary to pick marbles from EVERY urn.
I don't think I can make it in less than 5 moves (start by picking 6, then 4, then 3, then 2 and 1) but I am fairly sure it can be done in 4 or maybe less.
Any ideas?
combinatorics
asked Nov 27 at 14:12
Reyansh Laghari
1616
add a comment |
We have 15 urns each of them having a different number of marbles, from 1 to 15. We start by picking the same number of marbles from each of the urns we choose. We repeat the process until we have picked all marbles. What is the minimum number of days we can finish picking all marbles? Just to clarify that it is not necessary to pick marbles from EVERY urn.
I don't think I can make it in less than 5 moves (start by picking 6, then 4, then 3, then 2 and 1) but I am fairly sure it can be done in 4 or maybe less.
Any ideas?
combinatorics
asked Nov 27 at 14:12
Reyansh Laghari
1616
What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17
add a comment |
We have 15 urns each of them having a different number of marbles, from 1 to 15. We start by picking the same number of marbles from each of the urns we choose. We repeat the process until we have picked all marbles. What is the minimum number of days we can finish picking all marbles? Just to clarify that it is not necessary to pick marbles from EVERY urn.
I don't think I can make it in less than 5 moves (start by picking 6, then 4, then 3, then 2 and 1) but I am fairly sure it can be done in 4 or maybe less.
Any ideas?
combinatorics
asked Nov 27 at 14:12
Reyansh Laghari
1616
We have 15 urns each of them having a different number of marbles, from 1 to 15. We start by picking the same number of marbles from each of the urns we choose. We repeat the process until we have picked all marbles. What is the minimum number of days we can finish picking all marbles? Just to clarify that it is not necessary to pick marbles from EVERY urn.
I don't think I can make it in less than 5 moves (start by picking 6, then 4, then 3, then 2 and 1) but I am fairly sure it can be done in 4 or maybe less.
Any ideas?
combinatorics
combinatorics
asked Nov 27 at 14:12
Reyansh Laghari
1616
asked Nov 27 at 14:12
Reyansh Laghari
1616
edited Nov 27 at 14:30
asked Nov 27 at 14:12
Reyansh Laghari
1616
asked Nov 27 at 14:12
Reyansh Laghari
1616
asked Nov 27 at 14:12
Reyansh Laghari
1616
1616
What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17
add a comment |
What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17
What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17
What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17
add a comment |
2 Answers
2
active
oldest
votes
You can look at your urns as an array of 4 bit integers:
$0001_b$
$0010_b$
$0011_b$
...
$1111_b$
On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.
In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.
answered Nov 27 at 15:19
Rchn
49015
add a comment |
It is possible in 4 days:
First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.
Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.
Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.
Last day you took balls from all the nonemty urns.
answered Nov 27 at 15:19
greedoid
37.8k114794
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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%2fmath.stackexchange.com%2fquestions%2f3015813%2fpicking-marbles%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
You can look at your urns as an array of 4 bit integers:
$0001_b$
$0010_b$
$0011_b$
...
$1111_b$
On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.
In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.
answered Nov 27 at 15:19
Rchn
49015
add a comment |
You can look at your urns as an array of 4 bit integers:
$0001_b$
$0010_b$
$0011_b$
...
$1111_b$
On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.
In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.
answered Nov 27 at 15:19
Rchn
49015
add a comment |
You can look at your urns as an array of 4 bit integers:
$0001_b$
$0010_b$
$0011_b$
...
$1111_b$
On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.
In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.
answered Nov 27 at 15:19
Rchn
49015
You can look at your urns as an array of 4 bit integers:
$0001_b$
$0010_b$
$0011_b$
...
$1111_b$
On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.
In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.
answered Nov 27 at 15:19
Rchn
49015
edited Nov 27 at 15:24
answered Nov 27 at 15:19
Rchn
49015
answered Nov 27 at 15:19
Rchn
49015
answered Nov 27 at 15:19
Rchn
49015
49015
add a comment |
add a comment |
It is possible in 4 days:
First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.
Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.
Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.
Last day you took balls from all the nonemty urns.
answered Nov 27 at 15:19
greedoid
37.8k114794
add a comment |
It is possible in 4 days:
First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.
Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.
Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.
Last day you took balls from all the nonemty urns.
answered Nov 27 at 15:19
greedoid
37.8k114794
add a comment |
It is possible in 4 days:
First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.
Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.
Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.
Last day you took balls from all the nonemty urns.
answered Nov 27 at 15:19
greedoid
37.8k114794
It is possible in 4 days:
First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.
Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.
Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.
Last day you took balls from all the nonemty urns.
answered Nov 27 at 15:19
greedoid
37.8k114794
answered Nov 27 at 15:19
greedoid
37.8k114794
answered Nov 27 at 15:19
greedoid
37.8k114794
answered Nov 27 at 15:19
greedoid
37.8k114794
37.8k114794
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2fmath.stackexchange.com%2fquestions%2f3015813%2fpicking-marbles%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
What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17