Number Properties- Consecutive Integers #11

[Guest]

I can't seem to understand #11 in the Consecutive Integers chapter. I don't understand how the Factor Foundation Rule applies.

11. List six factors of the product of 5 consecutive even integers.

[StaceyKoprince]

It looks like you missed something in your transcription of the question - you just say "product of consecutive even integers" but you don't say how many consecutive even integers we're supposed to use. Please check the problem and post the complete wording. Thanks!

[Guest]

Sorry about that. 5 consecutive even integers.

[StaceyKoprince]

No problem! I've edited your original post so that it reads correctly now.

We're told that we have 5 consecutive even integers, but we're not told which 5. 2, 4, 6, 8, 10? 22, 24, 26, 28, 30? Who knows?

So we have to figure out the minimum that we can tell based only upon knowing that there are 5 consecutive even integers.

First, all even integers are divisible by 2. I have five of these even integers, so I have five factors of 2 to play with.

When I have the prime factors of any particular number, I can use them to find all of the factors of that number. Here's an example: 12
Prime factors: 2, 2, 3 (that is, 2*2*3 = 12)
Factors I can make from the prime factors: 2, 3, 2*2 (or 4), 2*3 (or 6), 2*2*3 (or 12)
So, all factors of 12 are: 1, 2, 3, 4, 6, 12 (don't forget the 1 - that's always a factor of any number!)

Back to the problem: I don't know what my overall number is, but I do know that five of the prime factors are 2's:
2, 2, 2, 2, 2
Use these to find other factors: 2, 2*2 (or 4), 2*2*2 (or 8), 2*2*2*2 (or 16), 2*2*2*2*2 (or 32)... and 1!! (Don't forget the 1!)

You could also technically find other factors as well. For example, every other even integer will be divisible by 4, not just 2, so I could say that at least two of those 5 consecutive even integers have a factor of 4, not just 2. But I'd rather make things as simple as possible for myself, and the above is the simplest way to find 6 factors.