It’s really very simple. Before starting the game, player A adds up all coins in odd positions, and all coins in even positions, and sees whichever total is greater.
Since there’s an even number of coins, he can always make sure to take coins on a specific (odd/even) position.
For instance, say odd ones are greater. On first play, he has choice of coin #1 or #50. By choosing 1, player B has to choose between #2 and #50, thereby exposing an odd one no matter which one he chooses. And so the pattern continues...
It’s really very simple. Before starting the game, player A adds up all coins in odd positions, and all coins in even positions, and sees whichever total is greater.
Since there’s an even number of coins, he can always make sure to take coins on a specific (odd/even) position.
For instance, say odd ones are greater. On first play, he has choice of coin #1 or #50. By choosing 1, player B has to choose between #2 and #50, thereby exposing an odd one no matter which one he chooses. And so the pattern continues...
So simple
If you notice, my first answer did go in the even odd direction but I couldn’t figure it out all the way. Good one!
2 friends decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then, alternating, each player takes on their turn one of the two coins at the ends of the line and keeps it. They continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as their opponent.
Here is the winning strategy.
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Let's assign the coins numbers from 1 to 50.
Tally the worth of the coins which have odd numbers.
Tally the worth of the coins which have even numbers.
If the odd coins are worth more, take coin nr. 1.
If the even coins are worth more, take coin nr. 50.
If they are worth the same, take either (or develop a strategy that maximises your gains by switching, but this is not always possible it will depend on the situation).
From then on, always take coins from the same side as your opponent.
This way, you will collect at least as much as him.
In some situations, you can maximise your gains by switching from odd to even or vice-versa, but this has to be done carefully it could also lead you to loose the game...
Since there is an even number off coins in the beginning, you are free to choose whether you want odd or even. Your adversary, however, will not be able to choose...
If you take away nr. 50, he can either take 1 or 49, both odd.
If you take away nr. 1 he can either take 2 or 50, both even.
If you then always take from the same end as him, he will be constrained to always stay with the oddd (or even) numbers...
He will not be able to switch from odd to even or vice-versa, unless you switch him, by not taking from the same end as him...
549,176,320 is a multiple of the first digit 8
As is every column separately549,000,000 is a multiple of 8
176,000 is a multiple of 8, and 320 is too. Plus all digits are different. So this number is pretty unique but I wonder if this is the answer you’re looking for.
549,176,320 is a multiple of the first digit 8
As is every column separately549,000,000 is a multiple of 8
176,000 is a multiple of 8, and 320 is too. Plus all digits are different. So this number is pretty unique but I wonder if this is the answer you’re looking for.
Well, 1000 is a multiple of eight (2x2x2x5x5x5), so it's logical that anything that ends in 000 is a multiple of 8...
Well, 1000 is a multiple of eight (2x2x2x5x5x5), so it's logical that anything that ends in 000 is a multiple of 8...
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176 and 320 are multiples without the thousands, that’s what made me make the first column work already as well. Plus those 9 digits are a multiple too so you can skip the 1,000 answer. But, all this aside, I’d be surprised if this is the answer she’s looking for. She probably wants “more”
100 mathematicians live on an island, along with the island's leader. 60 have brown eyes, and 40 have blue eyes, but no-one knows his own color. One day, the leader of the island (who has green eyes)made an announcement as follows:
"I can see at least one person with blue eyes. Starting tonight, there will be a plane at 8 pm every night. Whoever discovers they have blue eyes, must leave the island that night."
There are no mirrors on the island, and it is considered to be in very poor taste to discuss the subject of eye color. Of course, everyone knows what everyone else’s eye color is.
What effect did that announcement have? How many people leave the island and when?
100 mathematicians live on an island, along with the island's leader. 60 have brown eyes, and 40 have blue eyes, but no-one knows his own color. One day, the leader of the island (who has green eyes)made an announcement as follows:
"I can see at least one person with blue eyes. Starting tonight, there will be a plane at 8 pm every night. Whoever discovers they have blue eyes, must leave the island that night."
There are no mirrors on the island, and it is considered to be in very poor taste to discuss the subject of eye color. Of course, everyone knows what everyone else’s eye color is.
What effect did that announcement have? How many people leave the island and when?
Dunno in which direction to think
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Why did the leader say he sees at least 1
8 pm is dark. Does that matter?
No mirrors on an island but perhaps they can use the water for a mirror?
Do they want to leave the island? Otherwise as long as they don’t discuss eye color they’re good! Everyone gets to stay- since they don’t know yet that their eye color is blue.
