Picking marbles
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7
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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
asked Nov 27 at 14:12
Reyansh Laghari
1616
add a comment |
up vote
7
down vote
favorite
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 |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
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
up vote
4
down vote
accepted
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 |
up vote
3
down vote
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
36.4k114592
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
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 |
up vote
4
down vote
accepted
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 |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
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 |
up vote
3
down vote
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
36.4k114592
add a comment |
up vote
3
down vote
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
36.4k114592
add a comment |
up vote
3
down vote
up vote
3
down vote
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
36.4k114592
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
36.4k114592
answered Nov 27 at 15:19
greedoid
36.4k114592
answered Nov 27 at 15:19
greedoid
36.4k114592
answered Nov 27 at 15:19
greedoid
36.4k114592
36.4k114592
add a comment |
add a comment |
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What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17