July/August 2006 CPE Puzzler
 | A Strong WillA father, in his will left all his money to his children in the following manner: $1000 to the first born and 1/10 of what then remains, then $2000 to the second born and 1/10 of what then remains, then $3000 to the third born and 1/10 of what then remains, and so on. When this was done each child had the same amount. How many children were there? |
Answer: 0, 1, or 9 children.
Let x be the original amount.
Let y be the amount each child gets.
The first child got y = 1000+(x-1000)/10.
The second child got y = 2000+(x-2000-y)/10.
Here, we have two equations with two unknowns. This is enough to solve for x and y.
Using the substitution method,
1000+(x-1000)/10 = 2000+(x-2000-(1000+(x-1000)/10))/10
1000+x/10-100 = 2000+x/10-200-(100+(x/10-100)/10)
1000+x/10-100 = 2000+x/10-200-100-x/100+10
x/100 = 1000-200+10
x/100 = 810
x = 81,000
The amount each child gets can be calculated this way:
y = 1000 + (x-1000)/10 = 1000 + (81000 - 1000)/10 = 9000
The number of children is the original amount of money divided by the amount of money each child gets. That's 81,000 divided by 9,000, which is 9. So there are nine children.
Now, the tricky puzzle:
A father in his will left all his money to his children in the following manner:
The surviving children are paid in the order they were born. Before any money is handed out, the children "count off" in this order, so that they will be assigned an ordinal number. They are paid $1000 times the ordinal number plus 1/10 of the amount that remains in the estate. When this was done, all the children had the same amount. How many children were there?
You can solve this puzzle the same way as most people solved the first puzzle – by equating the first and second children's inheritance. But then you must also consider the case of just one child, who would receive $1000, with nothing left over. (The solver who pointed out this tricky solution equated the first and last children's inheritance, and got a quadratic that yielded the two solutions: 1 child or 9 children – very neat.)
I would argue that there is a third solution to the puzzle when phrased this way: zero children. If the father died broke and childless, then in a vacuous sense, all the father's money was left to his children, all the children received an inheritance, and they all got the same amount. (I stayed away from using the word "each", because, like "some", it can be argued that "each" implies at least one.)
Source: www.theproblemsite.com
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