Question

4329 = ___

• A. $(9\times {10}^0)+(2\times {10}^1)+(3\times {10}^2)+(4\times {10}^3)$
• B. $(4\times {10}^{-3})+(3\times {10}^1)+(2\times {10}^2)+(9\times {10}^3)$
• C. $(4\times {10}^{-3})+(3\times {10}^{-2})+(2\times {10}^{-1})+(9\times {10}^0)$
• D. $(9\times {10}^{-3})+(2\times {10}^{-2})+(3\times {10}^{-1})+(4\times {10}^3)$

Answer 1

The correct option is A

$(9\times {10}^0)+(2\times {10}^1)+(3\times {10}^2)+(4\times {10}^3)$

We know that $a^0=1$ for any non-zero integer $a.$
$4329=4000+300+20+9$
$=(4\times {10}^3)+(3\times {10}^2)+(2\times {10}^1)+(9\times {10}^0)$