Problem 2:
One coin is labeled with the number $1$, two different coins are labeled with the number $2$, three different coins are labeled with the number $3$,...,forty-nine different coins are labeled with the number $49$, and ffty different coins are labeled with the number $50$. All of these coins are then put into a black bag. The coins are then randomly drawn one by one. We need $10$ coins of any type. What is the minimum number of coins that must be drawn to make sure that we have at least $10$ coins of one type?
BdMO National Higher Secondary 2008/2
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Re: BdMO National Higher Secondary 2008/2
we have to pick at least 11476 coins...
lets pick 9 coins from each type (9 to 50) = 9+10+11+.......+50 =11475-36
then pick all of the coins from (1 to 8) = 1+2+3+......+8 = 36
total = 11475
so, we have pick 11476 coins to have at least 10 coins of each type.
lets pick 9 coins from each type (9 to 50) = 9+10+11+.......+50 =11475-36
then pick all of the coins from (1 to 8) = 1+2+3+......+8 = 36
total = 11475
so, we have pick 11476 coins to have at least 10 coins of each type.
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samiul_samin
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Re: BdMO National Higher Secondary 2008/2
the answer is 415. Because if we pick 9 coins of 9 to 50, we can pick 42*9=378 and 378+36=414.So, we we have to pick 415 coins to make sure that we have at least 10 coins of each type.☺☺☺