Hi Milan,

You are correct that one must be even, but one also must be a multiple of three since every third number is a multiple of three. Therefore I and III are correct. However, the product will only be a multiple of four if the group either includes a multiple of four (12 or 16), or starts and ends with an even number. So two groups, 13, 14, 15 and 17, 18 19 will be divisible by 2 and three, but not by four.

Tani, thank you again!

I did not know that the product of every third consecutive number must be a multiple of 3. Are there any other rules similar to that one?

Also, just becuase W is definately a multiple of three, how could I be sure that it is also a multiple of 6 (for example, 33 is a multiple of 3 yet 6 is not). If you can, please let me know where I am wrong.

It works for every integer - starting with 2 - every other number is divisible by two
starting with three - every third number is divisible by three
starting with 4.... etc.

starting with 17, every seventeenth number is divisible by 17 and on and on...

That's because to get multiples of a number you simply keep adding that number to the prior multiple