Discussion

Explanation:

$\frac{6^{73}}{4^{32}}=\frac{2^{73}\times 3^{73}}{{\left(2^2\right)}^{32}}=\frac{2^{73}\times 3^{73}}{2^{64}}=\frac{2^9\times 3^{73}}{1}$

Last digit of 2⁹ ≡ last digit of 2¹ = 2.

Last digit of 3⁷³ ≡ last digit of 3¹ = 3.

So, the required answer is the last digit of 2 × 3 = 6.

Hence, option (c).

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