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📅 May 19, 2026 👤 scraface

May 19, 2026

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📌 Thus, there are 14 integers among the first 100 positive integers that are congruent to 3 modulo 7. Therefore, the final answer is:

📌 \boxed{14}

📌 What is the largest integer that must divide the product of any three consecutive integers?

📌 Consider three consecutive integers denoted as \( n, n+1, n+2 \).

📌 Among any three consecutive integers, one must be divisible by 2 (since every second integer is even).

📌 Among any three consecutive integers, one must be divisible by 3 (since every third integer is a multiple of 3).

📌 Thus, the product \( n(n+1)(n+2) \) is divisible by both 2 and 3. Therefore, it is divisible by:

📌 Additionally, consider the divisibility by 4. Among any three consecutive integers, at least one is even, and at least one of these even numbers is divisible by 2, making the product divisible by \( 2 \times 2 = 4 \).

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