Choose the modification (change a specified operator and interchange the indicated digits) to make (6 ÷ 2) × 3 = 0 a true statement

Introduction / Context:
We need to turn a false equality into a true one by (i) changing one operator as indicated and (ii) interchanging a named pair of digits. Multi-edit puzzles like this demand precise sequencing and careful arithmetic.

Given Data / Assumptions:

• Initial statement: (6 ÷ 2) × 3 = 0 (false).
• Options indicate which operator to change and which digits to interchange in the LHS.
• Standard precedence and parentheses are observed after edits.

Concept / Approach:
Apply the operator change first, then interchange the digits throughout the LHS. Finally, evaluate and compare against the RHS (0).

Step-by-Step Solution:
Option C: change “×” to “−”; swap digits 2 and 3.After operator change: (6 ÷ 2) − 3.Swap 2 ↔ 3 in the LHS: (6 ÷ 3) − 2.Evaluate: 6 ÷ 3 = 2; then 2 − 2 = 0 ⇒ matches the RHS.

Verification / Alternative check:
Trying other options shows they yield nonzero results (e.g., swapping ÷ and × typically produces a positive number).

Why Other Options Are Wrong:
They leave a residual positive value because division and multiplication cannot be arranged to cancel to zero without the strategic subtraction obtained here.

Common Pitfalls:
Forgetting the digit swap after the operator change, or altering the content inside parentheses incorrectly.

Final Answer:
x to -, 2 and 3 transforms the LHS to 0, making the equation true.