Picking marbles
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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
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add a comment |
$begingroup$
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
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What do you mean by picking 6?
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– Akash Roy
Nov 27 '18 at 14:17
add a comment |
$begingroup$
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
$endgroup$
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
edited Nov 27 '18 at 14:30
Reyansh Laghari
asked Nov 27 '18 at 14:12
Reyansh LaghariReyansh Laghari
1666
1666
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What do you mean by picking 6?
$endgroup$
– Akash Roy
Nov 27 '18 at 14:17
add a comment |
$begingroup$
What do you mean by picking 6?
$endgroup$
– Akash Roy
Nov 27 '18 at 14:17
$begingroup$
What do you mean by picking 6?
$endgroup$
– Akash Roy
Nov 27 '18 at 14:17
$begingroup$
What do you mean by picking 6?
$endgroup$
– Akash Roy
Nov 27 '18 at 14:17
add a comment |
2 Answers
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$begingroup$
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.
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add a comment |
$begingroup$
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.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Nov 27 '18 at 15:24
answered Nov 27 '18 at 15:19
RchnRchn
49015
49015
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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 '18 at 15:19
greedoidgreedoid
46k1160117
46k1160117
add a comment |
add a comment |
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$begingroup$
What do you mean by picking 6?
$endgroup$
– Akash Roy
Nov 27 '18 at 14:17