Near-1000 Multiplication — Complement Method Evaluate 9998 × 999 efficiently by using (1000 − 2) × (1000 − 1).

Introduction / Context:
Numbers close to a power of 10 are perfect for the complement method. Computing 9998 × 999 is much easier if we rewrite both factors relative to 1000 and expand carefully, avoiding long multiplication.

Given Data / Assumptions:

• 9998 = 1000 − 2.
• 999 = 1000 − 1.
• We want (1000 − 2) × (1000 − 1).

Concept / Approach:
Use distributive expansion: (a − b)(a − c) = a^2 − a(b + c) + bc. Let a = 1000, b = 2, c = 1. This keeps arithmetic simple and exact while leveraging place-value structure.

Step-by-Step Solution:
1) Compute a^2 = 1000^2 = 1,000,000.2) Compute a(b + c) = 1000 × (2 + 1) = 3000.3) Compute bc = 2 × 1 = 2.4) Combine: 1,000,000 − 3000 + 2 = 997,002? Wait—this is for (1000 − 2)(1000 − 3). Re-evaluate directly: (1000 − 2)(1000 − 1) = 1,000,000 − 1000 − 2000 + 2 = 997,002? That still seems off—let’s expand directly:(1000 − 2) × (1000 − 1) = 1000×1000 − 1000 − 2000 + 2 = 1,000,000 − 3000 + 2 = 997,002. But our original factors are 9998 and 999 (not 998 and 997). Recognize a slip: 9998 = 10000 − 2, not 1000 − 2. Correct expansion follows.5) Correct form: 9998 × 999 = (10000 − 2) × (1000 − 1) = 10000×1000 − 10000 − 2000 + 2 = 10,000,000 − 12,000 + 2 = 9,988,002.

Verification / Alternative check:
Use 9998 × 999 = 9998 × (1000 − 1) = 9,998,000 − 9,998 = 9,988,002, confirming the corrected computation.

Why Other Options Are Wrong:
9997001 and 9987012 reflect misapplied complements; 9898012 is far off due to place-value errors; 9990002 ignores the subtraction step.

Common Pitfalls:
Confusing 9998 with 1000 − 2 instead of 10000 − 2; dropping thousands when subtracting; mixing terms in the distributive expansion.

Final Answer:
9988002