(\text{What approximate value will come in place of question mark (?) in the following questions?})( (82)^{8} \div (735)^{8} \times (731)^{6} \times 81 = 730^{?} )

Given: (82)^8 ÷ (735)^8 × (731)^6 × 81 = 730^?. This can be rewritten as: (82/735)^8 × (731)^6 × 81 = 730^?. Notice that 82/735 ≈ 1/9 (since 735 ÷ 82 ≈ 9), 731 ≈ 730, and 81 = 3^4. So: (1/9)^8 × 730^6 × 3^4 = 730^?. Simplifying: (1/9)^8 × 3^4 = (1/3^2)^8 × 3^4 = 1/3^16 × 3^4 = 1/3^12. But we need to express this in terms of base 730. Actually, let me think differently. Since 82/735 ≈ 1/9 = 3^(-2), we have: (3^(-2))^8 × 730^6 × 3^4 = 730^? 3^(-16) × 730^6 × 3^4 = 730^? 730^6 × 3^(-12) = 730^?. This doesn't simplify nicely to 730^4. Let me reconsider. Perhaps the question means: 82^8 ÷ 735^8 × 731^6 × 81 = (82/735)^8 × 731^6 × 81. If 82/735 ≈ 1/9, then (1/9)^8 × 730^6 × 81 = 3^(-16) × 730^6 × 3^4 = 730^6 × 3^(-12). For this to be a power of 730, the 3^(-12) must somehow cancel or convert. Actually, if 3^(-12) ≈ 730^(-2) (since 730^2 ≈ 532900, which is not close to 3^12 = 531441, they're very close!), then 730^6 × 730^(-2) = 730^4. Yes! 3^12 = 531441 and 730^2 = 532900, which are very close. So 3^(-12) ≈ 730^(-2). Therefore: 730^6 × 730^(-2) = 730^4. So ? = 4, which is option D.