Question : Solve the following.
$\left[25^2+8 \div 2^3-\left\{16+\left(28\right.\right.\right.$ of $\left.7 \div 2^2\right)-\left(18^2 \div 12^2\right.$ of $\left.\left.\left.\frac{1}{8}\right)\right\}\right]$
Option 1: 626
Option 2: 529
Option 3: 721
Option 4: 579
Correct Answer: 579
Solution :
$\left[25^2+8 \div 2^3-\left\{16+\left(28\right.\right.\right.$ of $\left.7 \div 2^2\right)-\left(18^2 \div 12^2\right.$ of $\left.\left.\left.\frac{1}{8}\right)\right\}\right]$
$=[625+8\div2^3-\{16+(196\div4)-(324\div18)\}]$
$=[625+8\div8-\{16+49-18\}]$
$=[625+8\div8-47]$
$=[625+1-47]$
$=579$
Hence, the correct answer is 579.
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