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Home » You’ll NEVER Fall Victim Again—Discover Norton Lifelock’s Surprising Protection! - Crosslake

You’ll NEVER Fall Victim Again—Discover Norton Lifelock’s Surprising Protection! - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Question: What is the least common multiple of the number of eggs laid by three different snake species, which are 12, 18, and 30?

📌 Solution: Factorize each number: $ 12 = 2^2 \times 3 $, $ 18 = 2 \times 3^2 $, and $ 30 = 2 \times 3 \times 5 $. The LCM is the product of the highest powers of all primes: $ 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180 $.

📌 \boxed{180}

📌 Question: What two-digit number is one more than a multiple of 9 and also one more than a multiple of 11?

📌 Solution: Let $ x \equiv 1 \pmod{9} $ and $ x \equiv 1 \pmod{11} $. This implies $ x \equiv 1 \pmod{99} $, since 9 and 11 are coprime. The two-digit numbers satisfying this are $ 99 + 1 = 100 $ (not two-digit) and $ 1 $. However, the next smaller solution is $ 1 $, which is not two-digit. Thus, no two-digit number satisfies both conditions.

📌 \boxed{1} (Note: The question may require reevaluation for valid constraints.)

📌 Question: What is the largest integer that must divide the product of any three consecutive integers in a population growth model?

📌 Solution: Among any three consecutive integers, there is at least one multiple of 2 and one multiple of 3. Thus, the product is divisible by $ 2 \times 3 = 6 $. For example, $ 1 \times 2 \times 3 = 6 $, $ 2 \times 3 \times 4 = 24 $, and $ 3 \times 4 \times 5 = 60 $, all divisible by 6. No higher integer (e.g., 12) divides all such products, as $ 1 \times 2 \times 3 = 6 $ is not divisible by 12.

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