A Strange will.
A father, in his will, left 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?
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1 solution
There are 9 children. Here are two solutions, working from either end of the line of children:
Let n be the number of children. 1) The last shall be first:
The last one got n * 1000 + 1/10 * 0 = a multiple of 1000. So they all got (the same) multiple of 1000 = n * 1000.
When the 2nd to last gets his share, there is 2 * n * 1000 available (since each gets the same, 2nd to last takes half of what's there). Last one's share = 2nd to last one's share n * 1000 = (n-1)* 1000 + 1/10 ( 2 * n * 1000 - (n-1)*1000)
Solve for n: n = n - 1 + (2 n - 1 n + 1) / 10 (dividing by 1000) 0 = -1 + (n+1) / 10 n+1 = 10 n = 9
and each child gets 9000, total 81000.
1000 + (81000-1000) / 10 = 1000 + 8000 = 9000 2000 + (72000-2000) / 10 = 2000 + 7000 = 9000 and so on 8000 + (18000-8000) / 10 = 8000 + 1000 = 9000 9000 + (9000-9000) / 10 = 9000 + 0 = 9000 2) Starting from the front of the line.
Each one gets 1/n of the "pot" = n * 1000. (see above for why that is the case).
Total pot = n * (n*1000) = n^2 * 1000.
First one gets n * 1000 = 1000 + (n^2 * 1000 - 1000) / 10
Divide by 1000: n = 1 + (n^2 - 1) / 10 10 n = 10 + n^2 - 1 n^2 - 10n + 9 = 0
(n - 9) (n - 1) = 0
n = 9 or n = 1.
The case n = 1 is not very interesting, or is precluded by the given, "to the second born ...".