Total number of positive integral factors of 70560 ?
Prime factorisation: \(70560 = 7056 \times 10 = 84^2 \times 10\)
\(84 = 2^2 \cdot 3 \cdot 7\), so \(84^2 = 2^4 \cdot 3^2 \cdot 7^2\). Then \(70560 = 2^5 \cdot 3^2 \cdot 5 \cdot 7^2\).
Number of factors = \((5+1)(2+1)(1+1)(2+1) = 6 \times 3 \times 2 \times 3 = 108\)
108 is more than 105 (but not more than 205). "More than 105" is the most precise true option.
Answer: (C) More than 105
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