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BMO question

Q: Matthew has a deck of 300 cards numbered 1 to 300. He takes cards
out of the deck one at a time, and places the selected cards in a row,
with each new card added at the right end of the row. Matthew must
arrange that, at all times, the mean of the numbers on the cards in
the row is an integer. If, at some point, there is no card remaining in
the deck which allows Matthew to continue, then he stops.
When Matthew has stopped, what is the smallest possible number of
cards that he could have placed in the row? Give an example of such
a row.

I know ive probably misintepreted the question, but why can't this min number of cards be two?? Since if you add one even and one odd number this would promise you that your mean would be not an integer

Reply 1

Original post
by yi123456
Q: Matthew has a deck of 300 cards numbered 1 to 300. He takes cards
out of the deck one at a time, and places the selected cards in a row,
with each new card added at the right end of the row. Matthew must
arrange that, at all times, the mean of the numbers on the cards in
the row is an integer. If, at some point, there is no card remaining in
the deck which allows Matthew to continue, then he stops.
When Matthew has stopped, what is the smallest possible number of
cards that he could have placed in the row? Give an example of such
a row.
I know ive probably misintepreted the question, but why can't this min number of cards be two?? Since if you add one even and one odd number this would promise you that your mean would be not an integer
If the first card was even, he could draw another even card.

So the question is after drawing the first card, can he :

then select another card so the sum is divisible by 2

then select another card so the sum is divislbe by 3

then select another card so the sum is divislbe by 4

....

You then want the smallest such selection such that its impossible to draw another card.
(edited 1 year ago)

Reply 2

Original post
by mqb2766
If the first card was even, he could draw another even card.
So the question is after drawing the first card, can he then:

then select another card so the sum is divisible by 2

then select another card so the sum is divislbe by 3

then select another card so the sum is divislbe by 4

....

You then want the smallest such selection such that its impossible to draw another card.

ohh that makes much more sense!! Thankss

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