Interchange exactly one pair of signs globally so that the equality holds true under normal precedence: 16 + 4 ÷ 2 − 21 × 7 = 21

Introduction / Context:
This is a sign-swap puzzle. We are given an arithmetic statement that is presently false and four options, each specifying a global swap of two operators. Exactly one swap, applied to every occurrence of the two signs, must convert the left-hand side (LHS) into a value equal to the right-hand side (RHS = 21).

Given Data / Assumptions:

• Expression: 16 + 4 ÷ 2 − 21 × 7.
• Target: equality with 21.
• Swaps are global (every instance is swapped), and standard precedence (×, ÷ before +, −) applies.

Concept / Approach:
Evaluate the LHS after applying each proposed global swap. The correct option is the one that yields 21 exactly. Compute carefully with operator precedence and avoid rounding errors.

Step-by-Step Solution:
Try Option D (swap × and ÷): LHS → 16 + 4 × 2 − 21 ÷ 7.Compute products/divisions first: 4 × 2 = 8; 21 ÷ 7 = 3.Now LHS = 16 + 8 − 3 = 21, which matches the RHS.

Verification / Alternative check:
Other swaps fail: e.g., “+ and −” gives 16 − 4 ÷ 2 + 21 × 7, which is far greater than 21; the remaining options similarly miss 21 when evaluated with precedence.

Why Other Options Are Wrong:
They produce outcomes that over- or undershoot 21 because they either amplify the large product term or fail to reduce it appropriately.

Common Pitfalls:
Forgetting that the swap is global; or ignoring precedence and evaluating left-to-right can lead to incorrect results.

Final Answer:
x and ÷ (swap multiplication and division globally) makes the equation true.