$\dfrac{{16 \times {{10}^2} \times 64}}{{{2^4} \times {4^2}}}$
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Ans: Use ${a^m} \times {a^n} = {\left( a \right)^{m + n}}$, ${a^m} \div {a^n} = {\left( a \right)^{m - n}}$ and ${a^{ - m}} = \dfrac{1}{{{a^m}}}$.
$\dfrac{{16 \times {{10}^2} \times 64}}{{{2^4} \times {4^2}}} = {\left( 4 \right)^2} \times {10^2} \times {2^{ - 4}} \times {4^3} \times {4^{ - 2}}$
= ${\left( {{2^2}} \right)^2} \times {10^2} \times {\left( {{2^2}} \right)^{ - 2}} \times {4^3} \times {4^{ - 2}}$
= ${4^2} \times {10^2} \times {4^{ - 2}} \times {4^3} \times {4^{ - 2}}$
= ${4^{2 - 2 + 3 - 2}} \times {10^2}$
= ${4^1} \times {10^2} = 400$
