If the integer n is greater than 1, is n equal to 2?

If the integer n is greater than 1, is n equal to 2?

(1) Any prime satisfies this statement (2, 3, 5, ...), thus n may or may not equal to 2. Not sufficient.

(2) n has at least following factors: 1 and n (the number itself). As given that "the difference of any two distinct positive factors of n is odd", then must be true that: \(n-1=odd\) --> \(n=odd+1=even\). Can \(n\) be even number more than 2? No, because if \(n=even>2\) it obviously has 2 as a factor and again as "the difference of any two distinct positive factors of n is odd", then n-2 must be odd, but \(n-2=even-2=even\neq{odd}\). Hence \(n=2\). Sufficient.

OR: "the difference of any two distinct positive factors of n is odd" means that number must have only one odd factor and only one even factor. (If odd factors, (or even factors) >1, then the difference of the pair of two odd factors (or even factors) will be even not odd). Only number to have only one odd and only one even factor is 2. Sufficient.

Answer: B.
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Statement (2): "the difference of ANY two distinct positive factors of n is odd", which means that when we pick ANY two distinct factors of n their difference must be odd.

So for n=20. Factors of n are: 1, 2, 4, 5, 10 and 20. 20-1=odd but 20-10=even, so n could not be 20.

Again: "the difference of ANY two distinct positive factors of n is odd" means that number must have only one odd factor and only one even factor. (If odd factors, (or even factors) >1, then the difference of the pair of two odd factors (or even factors) will be even not odd). Only number to have only one odd and only one even factor is 2.

Hope it's clear.
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Does two distinct factors mean that there are only two factors? If not, n can equal 20...5 and 4... 5-4=1. 1 is odd....

Oh..is it because if n=20, then 10 and 2 (which are also factors of 20) -- 10-2=8...even... not odd..

So does (2) mean that all factors of n that are different from each other, when subtracted, must be odd?

I don't have a solid understanding of the laws that would make this more concrete.

Bunuel, I have a doubt regarding statement 1. Can \(\sqrt{2}\) be a factor here? If so then statement 1 is sufficient to say that \(n\) is not equal to \(2\), because \(2\) has factors \(1, \sqrt{2}\) and \(2\).

Factor of an integer \(n\), is an integer which evenly divides \(n\) without leaving a remainder.

So \(\sqrt{2}\) is not a factor of any number.

Hope it helps.
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+1 B

Let's analyze statement (2):
The difference of any two distinct positive factors of n is odd.
Thisn means that any pair of factors of n is compound by an even number and an odd number.
This also means that n is a prime number. Why? Because if n had more than two factors, the difference of at least one pair of factors would be even. For example, if there were an additional odd number, there would be a pair of numbers whose difference would be even.
So, if n is a prime number whose has an odd factor and a even factor, then we are talking about the number 2.

ANSWER: B

I think I deserve kudos
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Hello Bunuel, Following the last post, I have the same doubt. Basically, if for example if I take number n = 6, then its two distinct factors 3 and 2 or factors 6 and 1 has the difference as odd... so n not equal to 2... can you please explain where am I wrong?

The difference of ANY two distinct positive factors of n is odd: if n=6 then 6-2=4=even, hence n cannot be 6, or any other number but 2.

Hope it's clear.
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Sir Bunuel,
I have a doubt here with regards to strategy:


(1) says : n has exactly two positive factors
This means that n is a prime no

(2) says: The difference between any two distinct positive factors is odd.

Now to plug in numbers to check the sufficiency of (2) , I didn't chose non prime numbers because (1) and (2) can never contradict each other.
So choosing 6,8 , 10 or any non prime no is basically stupidity. What do you say?

Regards,
Sach
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Yes, the statements in DS questions never contradict each other but your logic for testing numbers is not correct. Consider example below:

Is n=2?

(1) n is prime
(2) n is even

Now, if you test only even primes for the second statement you'll have that n could only be 2, since 2 is the only even prime. Therefore you'll get that (2) is sufficient, which is not.

Hope it's clear.
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Its not clear.
From what I understand
So if the 2 statements don;t contradict each other, I shuold still test numbers considering B as an individual statement without have any bearing of A.
Please confirm if my understnading is right.
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From F.S 1, we have that n has only two positive factors. Including the number itself and 1, this can only be a prime. Thus not sufficient as n can be any prime number.

From F.S 2, we have that the difference of any two positive factors is odd. We know that odd-even/even-odd = odd. So we know that "n" has at-least 2 factors , one of which is odd and one is even. Now we don't know whether this integer has only two factors. Say it has 3 factors. The third factor will either be odd or even. Now the presence of the statement any two distinct positive factors makes it essential that any two factors when subtracted give an odd integer. This will not be possible if we have any more factors than an odd factor and an even factor. Thus the number can only have only two factors, one odd and one even. This means the integer can only be 2. Thus, sufficient.

B.
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Yes. The two statements cannot contradict each other. But when analyzing one statement, you should as good as forget the previous one. (In some cases, one statement can give you an idea of what numbers you should try and hence be helpful but you will need to try others as well)

Take a simple example:

Is n divisible by 6?

1. n is even
2. n is a multiple of 3

When you try out statement 1, say, you try out 3 numbers: 2, 4, 6. You say 2 and 4 are not divisible by 6 but 6 is. So not sufficient.

When you try out statement 2, will you try only even multiples of 3? No. You will try all multiples of 3. 3 is not divisible by 6 but 6 is. Not sufficient. If you try only even multiples of 3, you will see that all even multiples of 3 are divisible by 6. So your answer will be 'sufficient'. But mind you, here you have used both statements together hence you will mark (C). So in essence, you did not analyze statement 2 alone at all. Answer could have been (B), we will never know (in the actual test!).

Hence, when analyzing each statement, do not look at the data of the other one. In fact, as far as possible, I try to re-read the question stem between the two statements to remind me of exactly what I have to consider and to help me forget the data I have already considered (else you might use it sub-consciously) Sometimes, one statement helps you cheat by giving you ideas of numbers you should try in addition to others, that's all.
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Maybe a silly doubt...When you say a number has 2 factors or 4 factors..is 1 also included in it?I usually exclude 1 when counting factors..pls cllarify..

1 is a factor of every integer, thus it's also included (the same way n is a factor of n, so n is also included).

Does this make sense?
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Hi Bunnel,

In this question , if the statement was " If n is an integer, is n equal to 2? " should we consider 0 as an integer in this case ?? is 0 an positive integer ?

Thanks,

0 is certainly an integer but it's neither positive nor negative.

Foe more Numper Properties tips check: number-properties-tips-and-hints-174996.html

Hope it helps.
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This one was very tough, but I got it right and could observe interesting number properties regarding this question.

(1) It says that n is a prime number. Clearly not sufficient
(2) Every odd > 1 number has ONLY odd factors, so the difference between factors will be always EVEN (ODD-ODD=EVEN)
Every even numer > 2 has both ODD and EVEN (# of even factors ≥ 2) factors, so their difference can be ODD and/or EVEN
And ONLY the difference of any two distinct positive factors of 2 is odd

Math experts please correct me if I'm wrong.

(1)
this is a disguised way of saying 'n is prime'
therefore, insufficient

(2)
this says any two factors. that means any two factors - i.e., ALL pairs of factors have an odd difference.
there's only one way to do this: one odd factor and one even factor. (as soon as you get 2 odd factors or 2 even factors, you get an even difference by subtracting them.)
for example take foll numbers:
factors of 3 = 1,3 3-1=even
factors of 4= 4,1,2 4-1=odd but 4-2=even
factors of 10 = 10, 1, 5, 2 10-1=odd; 10-2=even
so on and so forth
2 is the only # with only 1 odd factor and only 1 even factor.
therefore, sufficient
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