Final answer:
The statement is false because a number can have 2 and 4 as factors without having 8 as a factor. However, if a number has 8 as a factor, then it must have both 2 and 4 as factors due to the prime factorization of 8 being 2·3.
Step-by-step explanation:
The statement is false. If a number has 2 and 4 as factors, it does not necessarily have 8 as a factor; however, if a number has 8 as a factor, then it necessarily has both 2 and 4 as factors. Let's explore why this is the case with examples.
Consider the number 8. Since 8 is divisible by 2 and 4, it's true that any number that has 8 as a factor will also have 2 and 4 as factors. This is because 8=2·2·3, so if 8 is a factor, 2 and 2·2 (which is 4) must be factors as well.
However, a number can have 2 and 4 as factors without having 8 as a factor. For example, the number 12 has factors of 2 and 4 (since 12=2·6 and 12=3·4), but it is not divisible by 8. This is because having 2 and 4 as factors means the number must have at least two '2's in its prime factorization, but for 8 to be a factor, there must be three '2's, as 8=2×3.