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

Answer :

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

FRANCOCANACARIGO

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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