Option C\(a = 5! * X = 120 * X = 2^3 * 3^1 * 5^1 * X\)

\(a/2 = 2^2 * 3^1 * 5^1 * X\)

I: \(\frac{a}{2}\) - is not a divisor but a multiple of 10 - so NO

II: \(\frac{a}{2}\) - can be multiple of 8 or can't be, depends on value of X - so NO

III: \(\frac{a}{2}\) - can be written as \(2^1 * 3^1 * 10^1 * X\) - is definitely a multiple of 10 - so YES

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