Mixing
A game with numbered cards
Clash Royale CLAN TAG#URR8PPP
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Alice and Bob have each a number (not greater than 20) of cards numbered sequentially 1 onwards. Alice can easily reorder her cards and place them on a table one after the other so that no two cards which lie next to each other have a digit in common. On the other hand Bob, who has just one card more than Alice, has no way of placing his cards on a table likewise (no two cards lying next to each other with a shared digit), however he tries.
How many numbered cards does Alice have?
number-sequence
asked 1 hour ago
Bernardo Recamán Santos
1,7531034
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up vote
2
down vote
favorite
Alice and Bob have each a number (not greater than 20) of cards numbered sequentially 1 onwards. Alice can easily reorder her cards and place them on a table one after the other so that no two cards which lie next to each other have a digit in common. On the other hand Bob, who has just one card more than Alice, has no way of placing his cards on a table likewise (no two cards lying next to each other with a shared digit), however he tries.
How many numbered cards does Alice have?
number-sequence
asked 1 hour ago
Bernardo Recamán Santos
1,7531034
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Alice and Bob have each a number (not greater than 20) of cards numbered sequentially 1 onwards. Alice can easily reorder her cards and place them on a table one after the other so that no two cards which lie next to each other have a digit in common. On the other hand Bob, who has just one card more than Alice, has no way of placing his cards on a table likewise (no two cards lying next to each other with a shared digit), however he tries.
How many numbered cards does Alice have?
number-sequence
asked 1 hour ago
Bernardo Recamán Santos
1,7531034
Alice and Bob have each a number (not greater than 20) of cards numbered sequentially 1 onwards. Alice can easily reorder her cards and place them on a table one after the other so that no two cards which lie next to each other have a digit in common. On the other hand Bob, who has just one card more than Alice, has no way of placing his cards on a table likewise (no two cards lying next to each other with a shared digit), however he tries.
How many numbered cards does Alice have?
number-sequence
number-sequence
asked 1 hour ago
Bernardo Recamán Santos
1,7531034
asked 1 hour ago
Bernardo Recamán Santos
1,7531034
edited 17 mins ago
asked 1 hour ago
Bernardo Recamán Santos
1,7531034
asked 1 hour ago
Bernardo Recamán Santos
1,7531034
asked 1 hour ago
Bernardo Recamán Santos
1,7531034
1,7531034
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
I think that Alice must have
$17$ cards
Reasoning
In the first $18$ cards, that Bob will have, $10$ of those will contain the digit '$1$'. For any ordering, if we divide the cards into $9$ consecutive pairs, by the Pigeonhole Principle, there will be at least one pair which contain two cards showing the digit '$1$'. Hence, Bob will never be able to find a configuration that works.
On the other hand, Alice just needs to ensure all cards containing the digit '$1$' appear in the odd numbered positions and ensure no two cards which differ by $10$ are next to each other. For example, she could order them as follows, $$ 1, 2, 10, 3, 11, 4, 12, 5, 13, 6, 14, 7, 15, 8, 16, 9, 17$$
answered 30 mins ago
hexomino
31.9k295153
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I think that Alice must have
$17$ cards
Reasoning
In the first $18$ cards, that Bob will have, $10$ of those will contain the digit '$1$'. For any ordering, if we divide the cards into $9$ consecutive pairs, by the Pigeonhole Principle, there will be at least one pair which contain two cards showing the digit '$1$'. Hence, Bob will never be able to find a configuration that works.
On the other hand, Alice just needs to ensure all cards containing the digit '$1$' appear in the odd numbered positions and ensure no two cards which differ by $10$ are next to each other. For example, she could order them as follows, $$ 1, 2, 10, 3, 11, 4, 12, 5, 13, 6, 14, 7, 15, 8, 16, 9, 17$$
answered 30 mins ago
hexomino
31.9k295153
add a comment |Â
up vote
3
down vote
I think that Alice must have
$17$ cards
Reasoning
In the first $18$ cards, that Bob will have, $10$ of those will contain the digit '$1$'. For any ordering, if we divide the cards into $9$ consecutive pairs, by the Pigeonhole Principle, there will be at least one pair which contain two cards showing the digit '$1$'. Hence, Bob will never be able to find a configuration that works.
On the other hand, Alice just needs to ensure all cards containing the digit '$1$' appear in the odd numbered positions and ensure no two cards which differ by $10$ are next to each other. For example, she could order them as follows, $$ 1, 2, 10, 3, 11, 4, 12, 5, 13, 6, 14, 7, 15, 8, 16, 9, 17$$
answered 30 mins ago
hexomino
31.9k295153
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I think that Alice must have
$17$ cards
Reasoning
In the first $18$ cards, that Bob will have, $10$ of those will contain the digit '$1$'. For any ordering, if we divide the cards into $9$ consecutive pairs, by the Pigeonhole Principle, there will be at least one pair which contain two cards showing the digit '$1$'. Hence, Bob will never be able to find a configuration that works.
On the other hand, Alice just needs to ensure all cards containing the digit '$1$' appear in the odd numbered positions and ensure no two cards which differ by $10$ are next to each other. For example, she could order them as follows, $$ 1, 2, 10, 3, 11, 4, 12, 5, 13, 6, 14, 7, 15, 8, 16, 9, 17$$
answered 30 mins ago
hexomino
31.9k295153
I think that Alice must have
$17$ cards
Reasoning
In the first $18$ cards, that Bob will have, $10$ of those will contain the digit '$1$'. For any ordering, if we divide the cards into $9$ consecutive pairs, by the Pigeonhole Principle, there will be at least one pair which contain two cards showing the digit '$1$'. Hence, Bob will never be able to find a configuration that works.
On the other hand, Alice just needs to ensure all cards containing the digit '$1$' appear in the odd numbered positions and ensure no two cards which differ by $10$ are next to each other. For example, she could order them as follows, $$ 1, 2, 10, 3, 11, 4, 12, 5, 13, 6, 14, 7, 15, 8, 16, 9, 17$$
answered 30 mins ago
hexomino
31.9k295153
answered 30 mins ago
hexomino
31.9k295153
answered 30 mins ago
hexomino
31.9k295153
answered 30 mins ago
hexomino
31.9k295153
31.9k295153
add a comment |Â
add a comment |Â
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