Number Properties Chapter 4 No. 11 (4th Edition)

[Johnrodriguez05]

I understand how to do In Action Problem No. 11 in Chapter 4 of the Number Properties Guide; however, I do not not understand something that is stated in the explanation.

The explanation states that the number of different factors of (2^4)(3)=(4+1)(1+1)=10. It goes on to say that we learned this in chapter one. After reviewing chapter one, I still do not understand how they got to (4+1)(1+1). Any help would be greatly appreciated.

Thanks.

[jnelson0612]

John, that is a great question. I just went through the entire chapter plus In Action problems and explanations and cannot find that either. I have to assume this is an error. Thanks for bringing this up!

[thomaskorah]

Hi John,

I was on the same problem right now, and I couldn't figure it out either. It's clearly an error, but I googled to see if there is such a method, and I came across this, which explains how you can find the total number of factors for any given number:

http://www.gmathacks.com/gmat-math/numb ... teger.html

However, I was troubled by the following statement as well: "Furthermore, at least one of the integers must be divisible by 3, because there are three consecutive even numbers. Therefore the product will have a 3 as a prime factor."

There isn't anything in the material that makes it clear why. The material just says any 3 consecutive integers will be divisible by 3 (obviously), but why should it be the case for three consecutive EVEN integers?

Any help would be appreciated.
Thanks,
Thomas

[jnelson0612]

Hi John,

I was on the same problem right now, and I couldn't figure it out either. It's clearly an error, but I googled to see if there is such a method, and I came across this, which explains how you can find the total number of factors for any given number:

http://www.gmathacks.com/gmat-math/numb ... teger.html

However, I was troubled by the following statement as well: "Furthermore, at least one of the integers must be divisible by 3, because there are three consecutive even numbers. Therefore the product will have a 3 as a prime factor."

There isn't anything in the material that makes it clear why. The material just says any 3 consecutive integers will be divisible by 3 (obviously), but why should it be the case for three consecutive EVEN integers?

Any help would be appreciated.
Thanks,
Thomas

Hi Thomas,
Thanks for the link!

Okay, let's list out some consecutive even integers:
0*2*4
2*4*6
4*6*8
6*8*10
8*10*12

Do you notice that each of these sets of three contains a number divisible by 3? That is because all of these are multiples of 2. For example, to get 2*4*6, I take 2(1) * 2(2) * 2(3). To get 6*8*10, I take 2(3)*2(4)*2(5). So all consecutive multiples will also conform to the divisibility rule discussed earlier. Oh yes, and just because many students don't know this, zero IS divisible by all integers, so that first set listed yields a product which is divisible by 3.

Hope this helps!