List six factors of the product of 5 cons even intergers?

[cooper2248817]

this is from the number properties book.

List six factors of the product of 5 cons even intergers?



Can anyone expain? I don't understand the explanation from the book.

Thanks

[vgh101]

I'm assuming you're talking about #11 on pg. 45 of the MGMAT Number Properties book.

First, since the problem doesn't tell you what exact set of five consecutive even integers they're talking about, you list the five consecutive even integers as a, b, c, d, and e and make a prime box for each.

Now what do we know about a, b, c, d, and e? Well, the problem says they are consecutive EVEN integers. When you see "even" you think "divisible by 2". That means whatever a, b, c, d, and e might be, they will all DEFINITELY have at least one "2" in EACH of their prime boxes.

Here's where the Factor Foundation Rule from Chapter 1 comes into play: if these five consecutive even integers are multiplied together, and each has at least one 2 in each of their prime boxes, then we know that any combination of these five 2's will definitely be factors of the product abcde. There could be more than these, of course, but any combination of these five 2's will certainly be there.

So if the product abcde has five 2's in its prime box, then:

2 is a factor of abcde
2 x 2 = 4 is a factor of abcde
2 x 2 x 2 = 8 is a factor of abcde
2 x 2 x 2 x 2 = 16 is a factor of abcde
2 x 2 x 2 x 2 x 2 = 32 is a factor of abcde

But wait! That's only five factors. You need six. Ah ha...here's where you remember that 1 is always a factor of any number. Add 1 to your list, and you're done! An instructor can check this over to make sure it's sound or maybe have a quicker way to do it, but hope that helps.

[cooper2248817]

What is throwing me off is that 6 factors of a product of 5 cons even integers could be any number; why are we choosing: 1 and 5(2)s. Do we have to break it down to the basic primes? why can't we use five factors has 2, 4, 8, 16, 32 and 1.

how will this work for odd consecutive integers or this only works for even?

[vgh101]

Thankfully, based on the Factor Foundation Rule, whatever number abcde turns out to be it must have all the factors that a, b, c, d, and e have individually. Since, as we said earlier, any EVEN integer must have at least one 2 in its prime box, then any result of multiplying those 2's together MUST also be factors of abcde.

We're choosing 1 and five 2's because those are the prime factors we absolutely know MUST be factors of abcde. You're right, there could be any number of other factors. We're not saying the six we listed earlier are the ONLY ones; but whatever else is out there, those six will DEFINITELY be factors of abcde.

Think of it like a Data Sufficiency question. Sure, 128 might be a factor of abcde...but then again it might not. How can we know? Well we can't know because we aren't given any more information than that the numbers are (1) consecutive, (2) even, and (3) integers. For 128 to be a factor, we'd need to see seven 2's in our prime boxes...but the problem does not give us any information that enables us to just add two more 2's. We'd have to be able to justify adding more 2's, but nothing in the problem allows us to do so...therefore INSUFFICIENT to say 128 is a factor. Only go with the information you're given.

If the problem had instead said to list 6 factors of the product of five consecutive ODD integers, then we couldn't answer the question. This is because "odd" only means "not divisible by 2", and the only thing we'd know is that 2 will NOT appear in a, b, c, d, and e's prime boxes. But we would have no idea what WOULD go in the prime boxes, except 1. There would be no way we could definitively list five more factors. So the "even" part of the original question is key.

[StaceyKoprince]

Thanks for the explanations vgh. Very nice thinking and clear explanations. Yes, we could be talking about any 5 consecutive even integers, but certain things will be factors no matter what 5 consecutive even integers you choose.

Technically, we do know a little bit more, if we want to go there. We also know that every other even integer is divisible by 4. So either two or three of our 5 consecutive even integers are divisible by 4, not just 2! This would allow us to say that we have at least seven 2's for any set of 5 consecutive even integers. We might have more than that... we just don't know for sure.