A 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?
Choose the correct answer:
A) 9	B) 10	C) 11	D) 12	E) 13
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= 
<gadget id="calculator">1000 + (81000 - 1000)/10</gadget>
<output>9_000</output>
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.   9nswer 9
<result>A</result>
A 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?