Calculus problem with Summation I think?

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    Calculus Summation
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Peanuts fall into bucket 0 at a rate of one per second for 2^12 seconds, leading to a total of 4096 peanuts. Each bucket can hold up to four peanuts before the elephant takes action, consuming four and moving one peanut to the left bucket. This process continues until no peanuts remain to be eaten. The problem encourages exploration of the behavior of the system with smaller numbers, drawing a parallel to arithmetic in base 5. Ultimately, the discussion focuses on the distribution of peanuts across the buckets after all actions are completed.
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Homework Statement


A line of buckets numbered 0,1,2... extends indefinitely to the left with an elephant behind each bucket. Initially, all of the buckets are empty but then peanuts start falling into bucket 0 at a rate of one per second for 2^12 seconds. Whenever 5 peanuts accumulate in a bucket the elephant behind the bucket picks up all five, eats four, and places the remaining one in the bucket immediately to the left. When there are no more peanuts to be eaten, how many peanuts remain in each bucket?
 
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No, it's not a calculus problem. Try working it out for small numbers of peanuts. Doesn't it remind you of addition in arithmetic base 4?
 
Sorry, typo. Make that base 5.
 

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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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