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📰 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.
📰 Question: What is the remainder when the sum of the first 4 prime numbers after 20 is divided by 5?
📰 Solution: The first 4 primes after 20 are 23, 29, 31, and 37. Their sum is $ 23 + 29 + 31 + 37 = 120 $. Dividing 120 by 5 gives a remainder of $ 120 - 5 \times 24 = 0 $.
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