BrilliantBeard’s Game of Wits
You dig up a treasure chest full of 1 0 0 0 0 0 gold coins, but right when you are about to take it for yourself, you are ambushed by the infamous BrilliantBeard the Pirate and his nefarious crew of bandits!
Being prideful of his intelligence, BrilliantBeard says he will let you keep the treasure if you can beat him at a game of wits. In this game, you must alternate turns reducing the money in the chest to either 2 1 , 5 1 , or 1 0 1 of its previous amount, but splitting coins is not allowed. (For example, if the first person chooses to reduce the 1 0 0 0 0 0 gold coins to 1 0 1 of its previous amount, then there will be 1 0 0 0 0 coins left.) The first person that can reduce the money down to 1 gold coin is the winner and may keep the treasure chest. However, BrilliantBeard says that if you lose he will make you walk the plank!
Since you found the treasure, BrilliantBeard agrees to let you decide who will go first.
What should you choose?
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3 solutions
There is some ambiguity in the wording of this problem. I think it is just a quess as to what is meant in the problem as it is worded now.
firstly -your statement "you must alternate turns reducing the money in the chest by either1/2 ,1/5 or 1/10
, but splitting coins is not allowed." could easily be interpreted to mean that the coins in the box are to be reduced by 1/2, or 1/5 or 1/10 so the 100,000 reduced by 1/10 would be 90,000. I think that it would be more correct to say reduced to 1/10 of the remaining coins in the chest.
secondly- It is not clear what will happen if someone left with a number of coins that cannot be split evenly by 2,5,or 10. For example;suppose I left BrillinatBeard with 9 coins in the chest what are his options.
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Thanks, I added an example to hopefully resolve the ambiguity. For the second point, since 100,000 only has factors of 2, 5, and 10, it is not possible for 9 coins in the chest to be left over.
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It is possible for 5 coins to be left, though.
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@Martha Sapeta – Then the next person can reduce the 5 coins to 5 1 that amount to leave 1 coin left (and win).
I realize that 9 is not possible if the reductions are as you now have explained. But I actually think that the problem may be more interesting when I was thinking reduced by 1/10 as meaning leaving 9/10. In that case it is easy to arrive at many numbers that cannot be evenly divided. In that problem some type of clarification would be required to make a solution possible.
@David Vreken I started with 1/10 of 100,000 and on last turn( played by me) ,10 coins were left and I did 1/10 ( i.e. taken all 10 coins) , ZERO coins are left in the treasure chest and not ONE. Though 10/10 is one. I got the puzzle correct but I m still stuck on the problem mentioned above. Thanks.
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If you have 10 coins and choose to reduce it by 1 0 1 , there will be 1 0 ⋅ 1 0 1 = 1 coin left.
hahahahaha
Right, I'll gratefully slink away with my treasure chest and my one gold coin. And my life, I might add. :)
Another interesting thing about your solution is that I had not realized that after the initial move 1 0 0 0 0 0 → 1 0 0 0 0 you can just mirror BB's moves. Sometimes one doesn't see something that's staring at one in the face. I'll post my solution in a moment and you'll see what I mean... (if interested, obviously)
The Xs denote lost positions, while the circles denote winning positions, with an arrow indicating the winning move.
That's a good visual representation of the solution!
You should go first.
10^5=100,000, the original amount. After the treasure is divided by 10 five times, 1 coin will remain. Pick 1/10 first. No matter what BrilliantBeard chooses, you can pick a fraction to make sure his move times yours equals 1/10 or 1/100. (If he picks 1/10, pick 1/10 again. If he picks 1/2, you pick 1/5. If he picks 1/5, you pick 1/2.) Following this strategy, you are guaranteed a win.
I don't think this strategy always works. You pick 1/10 first and now there is 10,000 coins left. Then let's say BrilliantBeard chooses 1/2, so you choose 1/5, so now there are 1,000 coins left. Then let's say BrilliantBeard picks 1/10, so you pick 1/10, so now there are 10 coins left. Then BrilliantBeard can pick 1/10 and win.
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oh. well I feel dumb now lol
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It's not completely a wrong idea though, it would work if you could always make theirmove ⋅ yourmove = 1 0 1 . the issue is that when they do 1 0 1 and you do 1 0 1 then theirmove ⋅ yourmove = 1 0 0 1 instead of 1 0 1
Your strategy works if you choose 2 1 when they choose 2 1 and 5 1 when they choose 5 1 instead of 2 1 when they choose 5 1 and 5 1 when they choose 2 1 .
You should go first , and on your first turn reduce the treasure to 1 0 1 of its previous amount, so that there is 1 0 0 0 0 gold coins left. After this, you can copy BrilliantBeard's every move. (For example, if he chooses to reduce the treasure to 5 1 of its previous amount, then you should reduce the treasure to 5 1 of its previous amount on your next turn.) Doing this will guarantee that you will be the first to reduce the treasure to 1 gold coin and win.
After you won BrilliantBeard's game, he slammed the lid down on the 1 gold coin left in the treasure chest and sneered, "Aye, you’re pretty smart for a landlubber, but not quite smart enough for a pirate! As a man of my word you may keep the treasure chest as we agreed, but we never made an accord about the treasure. I’ll be keeping that for meself!”