Mixing
Initially there are m balls in one bag, and n in the other, where m, n > 0. Two different operations are...
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Initially there are m balls in one bag, and n in the other, where m, n > 0. Two different operations are allowed:
(a) Remove an equal number of balls from each bag;
(b) Double the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
Operation (b) is now replaced with (c):
(c) Triple the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
elementary-number-theory
asked Jan 19 at 17:41
John AburiJohn Aburi
142
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add a comment |
$begingroup$
Initially there are m balls in one bag, and n in the other, where m, n > 0. Two different operations are allowed:
(a) Remove an equal number of balls from each bag;
(b) Double the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
Operation (b) is now replaced with (c):
(c) Triple the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
elementary-number-theory
asked Jan 19 at 17:41
John AburiJohn Aburi
142
$endgroup$
add a comment |
$begingroup$
Initially there are m balls in one bag, and n in the other, where m, n > 0. Two different operations are allowed:
(a) Remove an equal number of balls from each bag;
(b) Double the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
Operation (b) is now replaced with (c):
(c) Triple the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
elementary-number-theory
asked Jan 19 at 17:41
John AburiJohn Aburi
142
$endgroup$
Initially there are m balls in one bag, and n in the other, where m, n > 0. Two different operations are allowed:
(a) Remove an equal number of balls from each bag;
(b) Double the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
Operation (b) is now replaced with (c):
(c) Triple the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
elementary-number-theory
elementary-number-theory
asked Jan 19 at 17:41
John AburiJohn Aburi
142
asked Jan 19 at 17:41
John AburiJohn Aburi
142
asked Jan 19 at 17:41
John AburiJohn Aburi
142
asked Jan 19 at 17:41
John AburiJohn Aburi
142
asked Jan 19 at 17:41
John AburiJohn Aburi
142
142
add a comment |
add a comment |
1 Answer
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(a),(b): Double the number of balls in the bag that contains the smaller number, until that number becomes at least as big as the number of balls in the other (if $m=n$, this is omittted). This is possible since $m,n > 0$.
We now have 2 bags with $x$ and $y$ balls, satisfying $xge y > x/2$. Take now $2y-x$ balls from each bag, which is a positive number and at most as big as the smaller number of balls in the bags, as $y ge 2y-x Leftrightarrow 0 ge y-x Leftrightarrow x ge y$.
This leaves $2x-2y$ balls in the bag that just had $x$ balls and $x-y$ balls in the bag that just had $y$ balls. Double the number of balls in the latter, then you have $2x-2y$ balls in both bags, and take them all out to get 2 empty bags.
(a),(c): Note that the total number of balls in both bags changes by an even amount for both operations (a) and (c). That means it is impossible to get to a total of $0$ balls in both bags from the starting position if $m+n$ is odd.
answered Jan 19 at 18:57
IngixIngix
4,502159
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1 Answer
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$begingroup$
(a),(b): Double the number of balls in the bag that contains the smaller number, until that number becomes at least as big as the number of balls in the other (if $m=n$, this is omittted). This is possible since $m,n > 0$.
We now have 2 bags with $x$ and $y$ balls, satisfying $xge y > x/2$. Take now $2y-x$ balls from each bag, which is a positive number and at most as big as the smaller number of balls in the bags, as $y ge 2y-x Leftrightarrow 0 ge y-x Leftrightarrow x ge y$.
This leaves $2x-2y$ balls in the bag that just had $x$ balls and $x-y$ balls in the bag that just had $y$ balls. Double the number of balls in the latter, then you have $2x-2y$ balls in both bags, and take them all out to get 2 empty bags.
(a),(c): Note that the total number of balls in both bags changes by an even amount for both operations (a) and (c). That means it is impossible to get to a total of $0$ balls in both bags from the starting position if $m+n$ is odd.
answered Jan 19 at 18:57
IngixIngix
4,502159
$endgroup$
add a comment |
$begingroup$
(a),(b): Double the number of balls in the bag that contains the smaller number, until that number becomes at least as big as the number of balls in the other (if $m=n$, this is omittted). This is possible since $m,n > 0$.
We now have 2 bags with $x$ and $y$ balls, satisfying $xge y > x/2$. Take now $2y-x$ balls from each bag, which is a positive number and at most as big as the smaller number of balls in the bags, as $y ge 2y-x Leftrightarrow 0 ge y-x Leftrightarrow x ge y$.
This leaves $2x-2y$ balls in the bag that just had $x$ balls and $x-y$ balls in the bag that just had $y$ balls. Double the number of balls in the latter, then you have $2x-2y$ balls in both bags, and take them all out to get 2 empty bags.
(a),(c): Note that the total number of balls in both bags changes by an even amount for both operations (a) and (c). That means it is impossible to get to a total of $0$ balls in both bags from the starting position if $m+n$ is odd.
answered Jan 19 at 18:57
IngixIngix
4,502159
$endgroup$
add a comment |
$begingroup$
(a),(b): Double the number of balls in the bag that contains the smaller number, until that number becomes at least as big as the number of balls in the other (if $m=n$, this is omittted). This is possible since $m,n > 0$.
We now have 2 bags with $x$ and $y$ balls, satisfying $xge y > x/2$. Take now $2y-x$ balls from each bag, which is a positive number and at most as big as the smaller number of balls in the bags, as $y ge 2y-x Leftrightarrow 0 ge y-x Leftrightarrow x ge y$.
This leaves $2x-2y$ balls in the bag that just had $x$ balls and $x-y$ balls in the bag that just had $y$ balls. Double the number of balls in the latter, then you have $2x-2y$ balls in both bags, and take them all out to get 2 empty bags.
(a),(c): Note that the total number of balls in both bags changes by an even amount for both operations (a) and (c). That means it is impossible to get to a total of $0$ balls in both bags from the starting position if $m+n$ is odd.
answered Jan 19 at 18:57
IngixIngix
4,502159
$endgroup$
(a),(b): Double the number of balls in the bag that contains the smaller number, until that number becomes at least as big as the number of balls in the other (if $m=n$, this is omittted). This is possible since $m,n > 0$.
We now have 2 bags with $x$ and $y$ balls, satisfying $xge y > x/2$. Take now $2y-x$ balls from each bag, which is a positive number and at most as big as the smaller number of balls in the bags, as $y ge 2y-x Leftrightarrow 0 ge y-x Leftrightarrow x ge y$.
This leaves $2x-2y$ balls in the bag that just had $x$ balls and $x-y$ balls in the bag that just had $y$ balls. Double the number of balls in the latter, then you have $2x-2y$ balls in both bags, and take them all out to get 2 empty bags.
(a),(c): Note that the total number of balls in both bags changes by an even amount for both operations (a) and (c). That means it is impossible to get to a total of $0$ balls in both bags from the starting position if $m+n$ is odd.
answered Jan 19 at 18:57
IngixIngix
4,502159
answered Jan 19 at 18:57
IngixIngix
4,502159
answered Jan 19 at 18:57
IngixIngix
4,502159
answered Jan 19 at 18:57
IngixIngix
4,502159
4,502159
add a comment |
add a comment |
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