Question: What is the fewest number of consecutive primes, starting with 2, that when summed produce a number divisible by 7?

Answer: The correct answer is 5.

The consecutive primes starting from 2 are

2, 3, 5, 7, 11

Summing gives 2 + 3 + 5 + 7 + 11 = 28

and 28 is divisible by 7

Thus the required number of primes is 5

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