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Sagot :
Answer:
5
Step-by-step explanation:
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
The fewest number of consecutive primes, starting with 2, that when summed produces a number divisible by 7 is 5.
To determine what is the fewest number of consecutive primes, starting with 2, that when summed produces a number divisible by 7, the following calculation must be performed:
- 2 + 3 = 5
- 2 + 3 + 5 = 10
- 2 + 3 + 5 + 7 = 17
- 2 + 3 + 5 + 7 + 11 = 28
- 28/7 = 4
Therefore, the fewest number of consecutive primes, starting with 2, that when summed produces a number divisible by 7 is 5.
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