Let $S$ be the number of trailing zeroes in $\left( \sqrt{2500i} +\sqrt{-2500i} \right)^{2500}$ , then find the sum of all the positive integral divisors of $S$ .

The answer is 9372.

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We use that $(1+i)^2 = 2i$ . Hence we begin to manipulate the given expression:

$(\sqrt{2500i} + \sqrt{-2500i})^{2500} = (\sqrt{2500i}(1+i))^{2500} = (2500i(2i))^{1250} = 5000^{1250}$

Now, it is obvious that there are $1250$ strings of $000$ trailing, ie there are $3 \times 1250 = 3750$ trailing zeroes. We see that $3750 = 2 \times 3 \times 5^4$ . Every divisor is of the form $5^n, 2 \times 5^n, 3 \times 5^n$ or $6 \times 5^n$ ; these forms are exhaustive and exclusive. Thus,

$\sigma_1(3750) = \sum_{n=0}^4 5^n + 2 \sum_{n=0}^4 5^n + 3 \sum_{n=0}^4 5^n + 6 \sum_{n=0}^4 5^n = 12 \sum_{n=0}^4 5^n = 12 \times \frac{5^5-1}{5-1} = \boxed{9372}$