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 marbles and a tenth of the remainder. The third child takes three marbles and a tenth of the remainder. And so on until the last child takes whatever is left. Knowing that all the children end up with the same number of marble, how many children were there and how many marbles did each one get?

Guest May 8, 2015

#3**+10 **

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 !!!

CPhill
May 8, 2015

#1**+10 **

LetÂ´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!

Guest May 8, 2015

#3**+10 **

Best Answer

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 !!!

CPhill
May 8, 2015