Verified answer

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 study!

Since they always take a tenth, they leave nine-tenths so there must be nine kids so the ninth kid takes 9 marbles leaving none

So when the eighth kid got there he took 8 + 1/10 of something of which 9/10 is 9, so there were 10 after he took the 8, so 18 when he got there.

When kid 7 got there, he took 7 + 1/10 of an amount such that 9/10 of it is 18, so 20 plus the 7 is 27 when he got there

When kid 6 got there he took 6 + 1/10 of 30 so 36

kid 5: 5 + 1/10 of 40 so 45

kid 4: 4 + 1/10 of 50 = 54

the totals are going up by 9 each time so kid 3 = 63, kid 2 = 72 and kid 1 = 81

And each kid gets 9 marbles

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