Mr. Norton, the eminent numismatist decides to divide his coin collection amongst his four children.

• The oldest child gets 1/2 of the collection,
• The second oldest child gets 1/3 of the collection.
• The third oldest child gets 1/n of the collection, where n is a positive integer.
• The youngest gets the remaining 50 coins.

How many coins at most can there be in the collection?

The total of the youngest two get:
C/n + 50 = C/6
6C + 50*6n = nC
nC - 300n - 6C = 0
n(C - 300) - 6(C - 300) = 1800
(n - 6)*(C - 300) = 1800
There are many integer pairs, (a,b) s.t. a*b = 1800
But we know n > 6 therefore C > 300, so a,b > 0

Maximizing C means maximizing (C-300), and minimizing (n-6).
If n=7 then (C - 300) = 1800

The maximum number of coins is
C = 2100

2100/2 = 1050
2100/3 =  700
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              2100

Edited on May 1, 2023, 9:33 am

Posted by Larry on 2023-05-01 09:32:02