Given that \frac{100\times99\times98\times\dots\times3\times2\times1}{100^{n}} is an integer. What is the <strong>LARGEST</strong> value of n for which this is true?

Explanation

We need the maximum power of 100 = 5^2 \times 2^2 dividing 100!. The exponent of 5 in 100! is \lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 24. To form 100^n = 5^{2n} \times 2^{2n} \dots, we must have 2n \leq 24, giving n \leq 12. Thus, none of the given values match.

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