This is a good one....!!!!!
Let N be the number of total marbles....
The first child recieves this many marbles 1 + 1/10 of the remainder =
1 + (N -1) / 10 = Equation (1)
The second child gets 2 plus 1/10 of what is left....
And what's left is: [ the N we started with minus the 1 marble the first child drew minus the 1/10 of what was left after that minus the 2 marbles the second child drew ] x 1/10...or, mathematically the second child gets....
2 + [N - 1 - (N-1)/ 10 - 2 ] / 10 = 2 + [ N - 3 - (N -1)/ 10 ] / 10 = Equation (2)
And since each child gets the same number of marbles....we can set (1) and (2) equal and solve
1 + (N-1)/ 10 = 2 + [N - 3 - (N-1) / 10] / 10
-1 + (N - 1) / 10 = [N - 3 - (N -1) / 10] / 10
-10 + (N -1) = [N - 3 - (N -1)/ 10]
-10 + N -1 = [10N - 30 - (N -1)] / 10
-100 + 10N - 10 = 10N - 30 - (N - 1)
-110 + 10N = 10N - 29 - N
-110 +29 = -N
-81 = -N
N = 81 and that's the number of marbles
So, using (1), each child gets 1 + (81-1)/ 10 = 9 marbles and since there are 81 of them....there must be 9 children !!!
