What is the sum of all the prime numbers between 35 and 70 which when divided by 12 leave a remainder that is a prime number?

Answer 1

There are 8 primes between 35 and 70: 37, 41, 43, 47, 53, 59, 61, 67

All prime numbers are 6n ± 1 since:

6n is divisible by 6

6n + 2 = 2(3n + 1) which is divisible by 2

6n + 3 = 3(2n + 1) which is divisible by 3

leaving only 6n + 1 or 6n + 5 = 6(n+1) - 6 + 5 = 6k - 1 (where k = n-1) as possible primes.

The remainder when one of these is divided by 12 is:

If n is odd, let n = 2m + 1

→ possible prime = 6(2m+1) ± 1 = 12m + 6 ± 1

→ possible primes are 12m + 5 or 12m + 7

So the remainder is 5 or 7 which are both primes when divided by 12.

If n is even, let n = 2m

→ possible primes = 6n ± 1 = 6(2m) ± 1 = 12m ± 1

→ possible primes are 12m + 1 or 12(m-1)+12 - 1 = 12k + 11 (where k = m - 1)

So the remainder is 1 or 11, of which only 11 is prime, when divided by 12.

Thus all primes which are not of the form 12m + 1 (= 6n + 1 when n is even) have a remainder which is prime when divided by 12.

if m = 3, 12m + 1 = 37

If m = 4, 12m + 1 = 49 = 7² (not prime)

If m = 5, 12m + 1 = 61

so the primes to sum are: 41, 43, 47, 53, 59, 67 which gives:

41 + 43 + 47 + 53 + 59 + 67 = 310

Answer 2

This question is based on ignorance or misunderstanding. The sum of all the primes between 35 and 70 is a number which is divisible by 12 and so leaves no remainder when divided by 12. That is, the remainder is 0 and, as you should know, 0 is not a prime number.