## 100 Cards, and others

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### 100 Cards, and others

Tried searching for various keywords and didn't see anything like this one. Fairly simple, but it's also fairly easy to tie yourself into knots trying to go down a wrong path, so I figured I'd post this one:
100 Cards with various visible values are laid out in a single file line. Two players will alternate. Each will be able to see all remaining cards during each turn, and must choose to pick up the card from either end. So the first player can pick the 1st or 100th, the second will be left to pick either the 2nd or 100th, or 1st or 99th, and so on. After each player has taken 50 cards (no passing, so the second player must pick up the last card, even if it is negative), the player with more total points (simply the sum of their cards' values) wins. In the case of a tie, the player who picked first wins.

How should the player going first assure victory?

---
Other quick ones:
You have 6 cards, showing "2", "3", "4", "5", "+", and "=". Use all 6 cards exactly once (face up, properly oriented, etc.) to make a true mathematical statement.

You are given 99 coins that are heads up, and an unknown number of coins that are tails up. You are blindfolded and somehow can't feel the difference between the heads and tails sides of the coins. You can count the coins, put them in arbitrary many piles, flip whichever coins you want, but remember, when you flip and when you sort, you DO NOT know which ones are heads up, which are tails up. In the end, you must end up with just two piles, each containing an equal number of heads. How do you do this?
BlueSoxSWJ

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### Re: 100 Cards, and others

BlueSoxSWJ wrote:Tried searching for various keywords and didn't see anything like this one. Fairly simple, but it's also fairly easy to tie yourself into knots trying to go down a wrong path, so I figured I'd post this one:
100 Cards with various visible values are laid out in a single file line. Two players will alternate. Each will be able to see all remaining cards during each turn, and must choose to pick up the card from either end. So the first player can pick the 1st or 100th, the second will be left to pick either the 2nd or 100th, or 1st or 99th, and so on. After each player has taken 50 cards (no passing, so the second player must pick up the last card, even if it is negative), the player with more total points (simply the sum of their cards' values) wins. In the case of a tie, the player who picked first wins.

How should the player going first assure victory?

Not too hard, given the tie resolution method.

Spoiler:
Add up the value of the odd indexed cards and the even indexed cards. If the sum of the evens is larger, always pick even, otherwise, always pick odd. You'll always be able to pick one of your preferred cards because you can pick one initially, and by doing so you force the opponent to pick one of the the unpreferred cards, and you'll be able to get one of the preferred cards next time because picking an unpreferred card reveals a preferred one.

BlueSoxSWJ wrote:Other quick ones:
You have 6 cards, showing "2", "3", "4", "5", "+", and "=". Use all 6 cards exactly once (face up, properly oriented, etc.) to make a true mathematical statement.

I think this one is new.

BlueSoxSWJ wrote:You are given 99 coins that are heads up, and an unknown number of coins that are tails up. You are blindfolded and somehow can't feel the difference between the heads and tails sides of the coins. You can count the coins, put them in arbitrary many piles, flip whichever coins you want, but remember, when you flip and when you sort, you DO NOT know which ones are heads up, which are tails up. In the end, you must end up with just two piles, each containing an equal number of heads. How do you do this?

This one has been done a few times.
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jestingrabbit

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### Re: 100 Cards, and others

BlueSoxSWJ wrote:You have 6 cards, showing "2", "3", "4", "5", "+", and "=". Use all 6 cards exactly once (face up, properly oriented, etc.) to make a true mathematical statement.

Spoiler:
3 ^ 2 = 4 + 5
Лом

Posts: 13
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### Re: 100 Cards, and others

BlueSoxSWJ wrote:You are given 99 coins that are heads up, and an unknown number of coins that are tails up. You are blindfolded and somehow can't feel the difference between the heads and tails sides of the coins. You can count the coins, put them in arbitrary many piles, flip whichever coins you want, but remember, when you flip and when you sort, you DO NOT know which ones are heads up, which are tails up. In the end, you must end up with just two piles, each containing an equal number of heads. How do you do this?

Spoiler:
Remove 99 coins, put them in a separate pile, then flip all of the 99 coins in this pile over. If the 99 coins originally consisted of h heads, after flipping there are 99-h heads. There are also 99-h heads in the other pile, since h heads were removed from an original 99.
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patzer

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### Re: 100 Cards, and others

Лом wrote:
BlueSoxSWJ wrote:You have 6 cards, showing "2", "3", "4", "5", "+", and "=". Use all 6 cards exactly once (face up, properly oriented, etc.) to make a true mathematical statement.

Spoiler:
3 ^ 2 = 4 + 5

Spoiler:
I don't think you are allowed to use an extra "^" card.
Please be gracious in judging my english. (I am not a native speaker/writer.)
lorb

Posts: 134
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Location: Austria

### Re: 100 Cards, and others

lorb wrote:
Лом wrote:
BlueSoxSWJ wrote:You have 6 cards, showing "2", "3", "4", "5", "+", and "=". Use all 6 cards exactly once (face up, properly oriented, etc.) to make a true mathematical statement.

Spoiler:
3 ^ 2 = 4 + 5

Spoiler:
I don't think you are allowed to use an extra "^" card.

Spoiler:
The 2 should be written as a superscript numeral rather than a "^", but I don't think it's possible to make superscript numerals on this site.
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patzer

Posts: 243
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### Re: 100 Cards, and others

INt€rN€t WaNd€r€R wrote:
lorb wrote:
Лом wrote:
BlueSoxSWJ wrote:You have 6 cards, showing "2", "3", "4", "5", "+", and "=". Use all 6 cards exactly once (face up, properly oriented, etc.) to make a true mathematical statement.

Spoiler:
3 ^ 2 = 4 + 5

Spoiler:
I don't think you are allowed to use an extra "^" card.

Spoiler:
The 2 should be written as a superscript numeral rather than a "^", but I don't think it's possible to make superscript numerals on this site.

Spoiler:
32 or 3^{2} or 3²

Or even 32, but that's getting a bit carried away.
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WarDaft

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### Re: 100 Cards, and others

The coins one is easy! Just take the pile of 99 heads and flip them: now both piles just contain tails, so they contain an equal number of heads!

What's that, you say? The 99 heads are all mixed in with the other coins?

Well, then you just...
Spoiler:
take out any 99 coins and flip them! (Suppose you took out N of the 99 heads: then the remaining pile contains 99-N heads. After flipping your pile, it also contains 99-N heads.)
mward

Posts: 53
Joined: Wed Jun 22, 2011 12:48 pm UTC

### Re: 100 Cards, and others

First post, so apologies if bumping oldish puzzles is not the right thing to do, but here are some other 'fun' answers to this puzzle:

Лом wrote:
BlueSoxSWJ wrote:You have 6 cards, showing "2", "3", "4", "5", "+", and "=". Use all 6 cards exactly once (face up, properly oriented, etc.) to make a true mathematical statement.

Spoiler:
3 ^ 2 = 4 + 5

Answers:

Spoiler:
If you're allowed to have different relative horizontal positions for different cards, as above, then:
Just have 2+3=5, with a 4 below it or far to the left. Then it's like you're naming this true statement "4".

Spoiler:
And if you're allowed to change other unspoken assumptions, then:
45+3 = 2
(Modulo 46...), And:
2+3=45
(Is a true mathematical statement giving part of a new definition for the operator "+" on the set {2,3,45}.

dudiobugtron

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