Difficulty: Easy
Correct Answer: 999999
Explanation:
Introduction / Context:This is an algebraic simplification and substitution problem. You substitute P = 99, compute P^2, evaluate the bracket, and then multiply. It also connects to the identity for (P + 1)^3 - 1, which often creates repeating 9 patterns.
Given Data / Assumptions:
Concept / Approach:Direct substitution works. Compute inside the parentheses first, then multiply by P. Keep arithmetic systematic to avoid mistakes with large numbers.
Step-by-Step Solution:
Compute P^2: 99^2 = 9801. Compute 3P: 3*99 = 297. Now bracket value: P^2 + 3P + 3 = 9801 + 297 + 3 = 10101. Multiply by P: P*(...) = 99 * 10101. Compute 10101*99 = 10101*100 - 10101 = 1,010,100 - 10,101 = 999,999.Verification / Alternative check:Notice that (P + 1)^3 = P^3 + 3P^2 + 3P + 1, so (P^3 + 3P^2 + 3P) = (P + 1)^3 - 1. Here P*(P^2 + 3P + 3) = P^3 + 3P^2 + 3P, which equals (P + 1)^3 - 1. With P = 99, (100)^3 - 1 = 1,000,000 - 1 = 999,999.
Why Other Options Are Wrong:
99999 and 9999 drop digits by mistakenly using 9 instead of 99. 9999999 adds an extra 9 due to multiplying by 100 instead of 99. 999909 is from incorrect addition inside the bracket (10101 must be exact).Common Pitfalls:Miscomputing 99^2, forgetting the +3 term, or doing 10101*99 as 10101*100 + 10101 instead of subtracting 10101.
Final Answer:999999
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