Solve ((2×3²+5)²÷(4²+3²-2))÷23: Complete Order of Operations Challenge

$\big((2\cdot3^2+5)^2:(4^2+3^2-2)\big):23=$

Let's simplify this expression while following the order of operations which states that exponents come before multiplication and division, which come before addition and subtraction, and that parentheses come before all of these,

Therefore, we'll start by simplifying the expressions in parentheses, note that in this expression there are parentheses within parentheses, where the first parentheses from the left are raised to a power and between the two pairs of parentheses there is a division operation, therefore, first we'll simplify the expressions in both pairs of parentheses, we'll do this according to the order of operations, first we'll calculate the values of the expressions with exponents (in both pairs of parentheses) then we'll calculate the result of multiplication in the first parentheses from the left and then we'll calculate the results of addition and subtraction operations in both pairs of parentheses:

$\big((2\cdot3^2+5)^2:(4^2+3^2-2)\big):23= \\ \big((2\cdot9+5)^2:(16+9-2)\big):23= \\ \big((18+5)^2:(16+9-2)\big):23=\\ \big(23^2:23\big):23$

Next we'll apply the exponent to the result of simplifying the expression in the first parentheses from the left, this is according to the aforementioned order of operations, and then we'll perform the mentioned division operation within the remaining large parentheses, finally - we'll perform the division operation that applies to the parentheses:

$\big(23^2:23\big):23 =\\ \big(529:23\big):23 =\\ 23:23=\\ 1$

Let's summarize the result of simplifying the expression, we got that:

$\big((2\cdot3^2+5)^2:(4^2+3^2-2)\big):23= \\ \big((2\cdot9+5)^2:(16+9-2)\big):23= \\ \big(23^2:23\big):23 =\\ 1$

Therefore the correct answer is answer A.

Note:

The final steps can of course be calculated numerically, step by step as described there, but note that we can also reach the same result without calculating the numerical value of the terms in the expression, by using the law of exponents for terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

We'll do this in the following way:
$\big(23^2:23\big):23 =\\ \frac{23^2}{23}:23=\\ 23^{2-1}:23=\\ 23:23=\\ 1$

First we converted the division operation in parentheses to a fraction, then we applied the aforementioned law of exponents while remembering that any number can be represented as itself to the power of 1 (and that any number to the power of 1 equals the number itself) and finally we remembered that dividing any number by itself will always give the result 1.