How many zeros are there in the product 1^{50}\times2^{49}\times3^{48}\times...\times50^{1}?

Explanation

Trailing zeros depend on the power of 5. The sum of exponents of multiples of 5 is \sum_{k=1}^{10} (51 - 5k) = 235. Multiples of 25 (25, 50) contribute an extra 26 + 1 = 27. Total powers of 5 = 235 + 27 = 262.

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