You begin with a single pile of 1,000 identical coins. You repeatedly do the following:

  1. Split any pile of size \(n\) into two parts of size \(x\) and \(n - x\),
  2. Add the product \(x(n - x)\) to a running total,
  3. Repeat until all piles are of size 1.

Question: What is the final total, and why is it always the same, no matter how you split?

Step 1: Understand the Process

Every time you split a pile of size \(n\) into two parts \(x\) and \(n - x\):

  • You generate a new term: \(x(n - x)\)
  • Eventually, you end with 1,000 piles of 1 coin each

So you make 999 splits total to go from 1 pile → 1,000 piles.

Step 2: Look for an Invariant

Let’s define a total function over all current piles:

\[ F = \sum_{i=1}^k n_i^2 \] where \(n_i\) is the size of each pile and \(k\) is the number of piles.

Observe what happens in one split:

  • You replace a pile of size \(n\) with two piles \(x\) and \(n - x\)
  • The total \(F\) changes from: \[ n^2 \to x^2 + (n - x)^2 = n^2 - 2x(n - x) \]

So each split reduces \(F\) by exactly: \[ \Delta F = -2x(n - x) \]

Hence, the term \(x(n - x)\) added to the running total corresponds to a drop of \( \Delta F = -2x(n - x) \).

So if \(T\) is the running total sum of all \(x(n - x)\), then:

\[ F_{\text{initial}} - F_{\text{final}} = 2T \]

Step 3: Apply the Invariant

  • Start with one pile of size 1,000 → \(F_{\text{initial}} = 1000^2 = 1,000,000\)
  • End with 1,000 piles of 1 → \(F_{\text{final}} = 1,000 \times 1^2 = 1,000\)

Thus:

\[ 1,000,000 - 1,000 = 2T \Rightarrow T = \frac{999,000}{2} = 499,500 \]

Final Answer

The final total is always 499,500.
It is invariant under the splitting choices because the decrease in the sum of squares \(F\) is always twice the accumulated total.

Reference