Question 631773: I can't figure this one out
Joe plays marbles with his friends and Joe finds out he is moving away. Joe has a large number of marbles that he decides to split among his friends. To his best friend, he gives 1 marble plus 1/7 of what is remaining. To his second best friend, he gives 2 marbles plus 1/7 of what is remaining. To his 3rd best friend, he gives 3 marbles plus 1/7 of what is remaining. He does this consecutively with all of his friends, until the last friend gets all that is remaining. After this is all done, everyone realizes they have the exact same number of marbles. How many marbles did Joe have and how many friends played marbles, including Joe?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! I found the wording a bit confusing. However, the two interpretations I considered gave me different numbers of marbles, but the same number of marble-playing friends.
As written, the second to the last friend (friend number x) gets x marbles plus 1/7 of the rest. If we call that 1/7 of the rest m, friend number x gets x+m marbles. The last friend (friend number x+1) will get the other 6/7 of that rest (6m), which would be the same amount, so 6m=x+m, and x=5m.That means Joe gave marbles to x+1=5m+1 friends. Because every one of those friends got the same 6m amount, we can calculate the total number of marbles Joe gave away as (5m+1) times (6m). Choosing m, we can calculate those numbers. For each total number of marbles, we can work out what the first, second, third, ... friends gets, and it should always be 6m. It only works for m=1. Then Joe gives 6 marbles to each of 6 friends for a total of 36 marbles given away. Counting Joe, there were  friends playing marbles together.
At first I thought that Joe was giving away his marbles to one friend at a time, giving each friend 1/7 of what he had at the moment plus the number representing the friend's order in the giveaway (1 extra marble to the first friend, 2 to the second one, and so on).
Warning: Long explanation follows.
In that case, if Joe has  friends.
Let his best friend be friend number 1,
his second best friend be friend number 2, and so on,
up to his least favorite friend that would be friend number .
By the time Joe gets to friend number he gives him the remaining  marbles, which are also 1/7 of the remaining marbles, plus .
So 
-->  (multiplying both sides times 7)
-->  (subtracting R from both sides)
Because and are positive integers,
must be a multiple of 7, and
must be a multiple of 6.
 with  can be proved to be a solution by just checking it.
Are there any other solutions?
We know that <-->  .
Let's find another equation.
Joe gave up all his marbles by giving marbles to each of his friends.
So, he must have had  marbles.
He gave 1/7 of that plus one more marble to friend number 1, and that was also marbles, so
Substituting , that turns into
 --> 
-->  (multiplying both sides times 7)
-->  , and that is our second equation,
which we can solve by factoring:
--> 
The solutions to that equation are and 
substituted into gives , which is very reasonable.
substituted into gives  , which is not reasonable because Joe would not cut any marbles into fractions.
The only solution that works for both equations is with
Joe had 42 marbles and gave 7 to each of 6 friends.
Counting Joe, the whole group of marble-playing kids is made of
friends.
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