A group of children share marbles from a bag. The first child takes one marble and a tenth of the remainder. The second child takes two marb

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 !!!

Let´s suppose that there are x marbles. 
Then we have the following steps: 
1st child: C1 = 1 + (x-1)/10 
2nd child: C2 = 2 + (x - C1 - 2)/10 
3rd child: C3 = 3 + (x - C1 - C2 - 3)/10 
... 
n-th child: Cn = n + (x - C1 - C2 - ... - C[n-1] - n)/10 

But C1 = C2 = C3 = ... = Cn = k (given by the problem) 

We can rewrite the equations in this way: 
1st child: k = 1 + (x-1)/10 
2nd child: k = 2 + (x - k - 2)/10 
3rd child: k = 3 + (x - 2*k - 3)/10 
... 
n-th child: k = n + (x - (n-1)*k - n)/10 

We must find n. If we choose any two equations, we can find k and x: 
k = 1 + (x-1)/10 
k = 2 + (x - k - 2)/10 
Solving the linear system we find: k=9 and x=81. 
If we try to use the last equation: k = n + (x - (n-1)*k - n)/10 to find the value of n, we will get 0 = 0. So we wont be able to find the value of n with this last equation. Well, we know we have 81 (value of x) marbles. And we know that each child has 9 marbles (value of k). Hence we have 9 children! 

Good job, Anonymous!

This does look like a great question with equally great answers.     

Thanks, Melody....I really liked this one , too!!!!!

Time is always a problem :(

There are so many great questions I want to do them all.

AND

There are so many great answers and I want to learn from all of those too.  :((