Manhattan Number Prop. chapter 2 concept not understood

[anuj_sharma63]

In chapter 2 one DS question says that If x>1 and there are x unique factor of x?
Answer explained is totally not understood.
What are uniques factors and why condition sets is valid for 1 and 2 and why not for 3 or 4.
Please explain as I have just started the preperation and I am struck.

Anuj

[jlucero]

Per forum rules, please post the full question and answer choices so that this question can be used by others as well.

As for your question, "unique factors" just mean different factors. The number 12 has 6 unique factors- 1, 2, 3, 4, 6, and 12. Let's take the first few numbers greater than 1 and see how many factors they have:

2- 2 factors (1, 2)
3- 2 factors (1, 3)
4- 3 factors (1, 2, 4)
5- 2 factors (1, 5)
6- 4 factors (1, 2, 3, 6)

Notice that as we choose larger numbers, they can have more factors, but they can never have ALL the factors from 1 to that number. So this statement is sufficient because if x has x factors, that number needs to be the same: 2 has 2 factors, but 3 doesn't have 3 factors, and 88 doesn't have 88 factors.

[anuj_sharma63]

but they can never have ALL the factors from 1 to that number:
I didnt understand this statement.PLease explain.
Second unique factors means there is no repeation of factors?
For e.g. 8 has 3 factors count but same number 2. So unique will be only 2 once.Correct me If I am wrong.

Anuj SHarma

[tim]

1) Joe explained this quite well in his post. Was there something about his explanation you didn't understand? If so, you'll have to be more specific.

2) No repetition. So for 8 we only count 2 once, but we also count 1 once, 4 once, and 8 once, giving us four unique factors for 8 (the numbers 1, 2, 4, and 8). Your assertion that "8 has 3 factors" is simply not true.

[anuj_sharma63]

factor which you have mentioned as in (1,2,4,8) are OK. but what I asserted was 8 has 2^3 factor hence three 2's.
If thats so then what I aseerted is called Prime Factors only.Please correct me if I am wrong.
This is important for me to understand the concept.

[tim]

Yeah I think you're confusing the concept of factors with a prime factorization. The prime factorization of 8 does indeed consist of three 2's, but the factors are 1,2,4,8.