NeedingWD40
New member
- Joined
- Jan 14, 2019
- Messages
- 18
Numbers that are the product of 2 primes have exactly 4 factors e.g 6 has 1, 2, 3, 6 sums to 12; 15 has 1, 3, 5, 15, sums to 24 etc
My question is, is the sum of those factors always a multiple of 3?
For the case where 2 is one of the primes then I think so:
If 2 is one of the prime factors of x, 2 and x must be even, 1 and the other prime factor must be odd (as 2 is the only even prime). We can express all odd numbers as two times a number plus 1 e.g. 2*a +1. So the factors of x are 1, 2, 2*a+1, 2(2*a+1) as x must be the product of 2 and the other factor. The sum of these is 1+2+2a+1+2(2a+1) = 1 + 2 + 2a + 1 + 4a + 2 = 6 + 6a = 3(2 + 2a) so is always a multiple of 3.
But when I tried this approach with two distinct odd primes e.g. 2a+1 and 2b+ 1 I end up with
1, 2a+1, 2b+ 1, (2a+1)(2b+1) which sum to 4 + 4ab + 4a + 4b... definitely a multiple of 4 but doesn't look like a multiple of 3...
My question is, is the sum of those factors always a multiple of 3?
For the case where 2 is one of the primes then I think so:
If 2 is one of the prime factors of x, 2 and x must be even, 1 and the other prime factor must be odd (as 2 is the only even prime). We can express all odd numbers as two times a number plus 1 e.g. 2*a +1. So the factors of x are 1, 2, 2*a+1, 2(2*a+1) as x must be the product of 2 and the other factor. The sum of these is 1+2+2a+1+2(2a+1) = 1 + 2 + 2a + 1 + 4a + 2 = 6 + 6a = 3(2 + 2a) so is always a multiple of 3.
But when I tried this approach with two distinct odd primes e.g. 2a+1 and 2b+ 1 I end up with
1, 2a+1, 2b+ 1, (2a+1)(2b+1) which sum to 4 + 4ab + 4a + 4b... definitely a multiple of 4 but doesn't look like a multiple of 3...