### Brain Teasers

# Coin Collection II

Another numismatist decides to divide his coin collection among his children. The oldest gets 1/2 of the collection, the second gets 1/3, the third gets 1/n, where n is a natural number, and the youngest gets the remaining 50 coins. How many coins at most can there be in the collection?

### Answer

There are 2100 coins in the collection and n=7.1050 + 700 + 300 + 50 = 2100

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## Comments

do u like math??

Yes, this is why I use digits as my username.

great teaser... increasing n above would decrease the total number of coins and an n lower than 7 would make the coin collection non-existent as 1/2+1/3+1/6 is already equal to 1...

Yes very nice teaser.

I'm sorry to say that I didn't see the simplicity of it and solved it by solving the equation

x = x/2 + x/3 + x/n + 50

which reduces to

x = 300n / (n-6)

At this point what should have been initially obvious became obvious: n has to be at least 7 to make the number of coins a positive number and the number of coins decreases as n increases, so

x = 300*7 / (7-6)

x = 2100 / 1

x = 2100

I'm sorry to say that I didn't see the simplicity of it and solved it by solving the equation

x = x/2 + x/3 + x/n + 50

which reduces to

x = 300n / (n-6)

At this point what should have been initially obvious became obvious: n has to be at least 7 to make the number of coins a positive number and the number of coins decreases as n increases, so

x = 300*7 / (7-6)

x = 2100 / 1

x = 2100

I like this teaser except for one thing. You asked how many coins there can be at most, to which the answer is indeed 2100. However, your answer as stated says that "there are" 2100 coins. There may not be that many in the collection. At least not how the question is worded.

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