The question is this: You know that a number is divisible by 6 if it is divisible by both 3 and 2. So why isn't a number divisible by 8 if it is divisible by both 4 and 2? My answer is this: The only prime factors of 6 are 2 and 3. So, if a number is divisible by both, then it is divisible by their product,6. The prime factors of 8 are 2, 3 times. In order for a number to be divisible by 8 it must have 3 factors of 2. If a number is divisible by 2 the quotient of that divisible by 4 , then the original number is divisible by 8. Is this right?

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Your explanation is almost correct, but there is a small error. The prime factors of 8 are actually only 2, raised to the power of 3 (not 2 raised to the power of 3 times). Therefore, in order for a number to be divisible by 8, it must have at least 3 factors of 2.

To determine whether a number is divisible by both 4 and 2, you need to consider their respective prime factors. Let's break it down step by step:

1. To check if a number is divisible by 4, it must have at least 2 factors of 2. The prime factorization of 4 is 2^2.
- For example, the number 16 is divisible by 4 because it can be written as 2^4 (which has 2 factors of 2).

2. To check if a number is divisible by 2, it must have at least 1 factor of 2.
- For example, the number 12 is divisible by 2 because it can be written as 2^2 * 3 (which has 1 factor of 2).

So, if a number is divisible by both 4 (2^2) and 2 (2^1), it means it has at least 2+1=3 factors of 2. Therefore, the number is divisible by 8 (2^3).

In summary, a number is divisible by 6 if it is divisible by both 3 and 2, and a number is divisible by 8 if it is divisible by both 2^2 (4) and 2^1 (2), having a total of at least 3 factors of 2.

no

correct, but wordy.

Six