If x is an integer greater than 1, what is the value of x?

Not sure I understand your question here, but we use the fact that x cannot be divisible by x-1 unless x=2 to proof that x can only be 2.

Need an explanation for statement 2.... the number to be added to x has to be a odd prime...for the result to be odd...but then why cannot x be 2 here.... I agree with the OA...it will be A...but I don't understand why can't x = 2 be one of the possible values....

From (2) x can be 2, but it can also be any other even number, for example 4: 4+5=9=odd.

Hope it's clear.

Thanks Bunuel... I think I might have got confused by the language of the earlier post......Thank you for the clarification

X > 1, what is x?

(1) if x=2, unique factor is 2
If x=3, unique factor is 2
Moving up the unqie factor count minimum is 2 and only 2 qualifies for the condition
X=2, SUFFICIENT

(2) x + odd = odd
We don't know x yet excep it is x>1.
so if prime is even, x should be odd
If prime is odd, x should be even.
INSUFFICIENT

Answer: A

Hi,
Pls answer below points:
1) x to have x distinct positive factors it must be divisible by EVERY integer from 1 to x- Not clear about this statement. As per my understanding factors are those which completely divide the number.So how can factors will be divisible and that too why from all the number b/w 1 to x.
2) What do you understand by 'unique factors'.

Please explain me the strategy to adopt while attempting to comprehend the concepts.

Regards
Anuj

Factor of x is a positive integer which divides x evenly (without a remainder). Notice that each factor of x must be less than or equal to x: the smallest factor of a positive integer is 1 and the greatest factor is that integer itself. For example, the smallest factor of 12 is 1 and the greatest factor of 12 is 12 itself.

Now, x to have x distinct positive factors it must be divisible by EVERY integer from 1 to x: 1, 2, 3, ..., x-1, x --> total of x factors.

As for your second question: unique factors are distinct factors. For example 12 has 6 unique (distinct) factors: 1, 2, 3, 4, 6, and 12.

Hope it's clear.

I have got your point completely.Thanks for giving your time!
one thing i need to get clarify is Why we are taking X-1 always instead for any series X+1 for consecutive integers is also possible or not?

Not sure understand your question: where are we taking x-1 consecutive integers? Can you please elaborate?

The unique factors part tricked me..

Hey,

When they say there are x unique factors of x try substituting values for x.
As x> 1 , suppose x= 5 There ARE NOT 5 unique factors of 5. (There are just 2:- 1,5) .Try substituting different numbers, u will realize that the only possible case is when
x=1 and x =2 .[1 has 1 unique factor(1), 2 has 2 unique factors (1,2)]
Question stem says that x > 1 , thus x= 2. Sufficient.

Statement 2 is insufficient as x could be 2/4/6 etc.

Answer :-A
Hope it helps.

Thank's Bunuel

Can someone please check my approach here?

The approach I tried this one is as follows -

Since it is given that x>1 and value of x has to be an integer.

St. A - the least possible unique number (between 0 to 9) from 2 unique factors will be 6 - (2,3)
Sufficient

St. B - (x + prime no.) > odd no. x
It can be (3+2) > 3
or can be (5+2) > 5.
Thus Insufficient.

Answer - A

Hello everyone/ Bunuel

Can anyone please explain statement 1 and correct my understanding here: when I try substituting values of x say 9 then x = (1,3,9), since the questions stem says that x>1, thus unique factor is 2. But when we have larger numbers say x=20 then x= (1,2,4,5,10,20) here the unique factors are 5 (when x>1). Hence how can x be 2 as per the above answers. Kindly explain, if I have any gap in my understanding.

Thanks

Statement 1

I) there are x unique factors of x

Only the prime number '2' has two factors 1, and 2. Hence, the statement is sufficient to answer the question. We can eliminate B, C, and E.

Statement 2

II) the sum of x and all prime factors greater than x is Odd

All prime numbers except 2 are odd and hence the sum of all prime factors greater than 2 should be even.

From the question premise, we know that x is an integer greater than 1, so the minimum value of x can be 2.

Let's assume x = 2, so 2 + even = even. Therefore x cannot be 2, or for that matter any even integer. It can however be any odd integer.

Say
x = 3 → 3 + (sum of all prime factors greater than 3) = 3 + even = odd
x = 5 → 5 + (sum of all prime factors greater than 5) = 5 + even = odd

As we do not have a unique value for x, the statement alone is not sufficient.

Option A


P.S. = IMO the verbiage of the question (esp St2) isn't great. We could have referred to them as prime integers instead of prime factors.


Statement 2

(2) The sum of x and any prime number larger than x is odd

All prime numbers except 2 are odd.

From the question premise, we know that x is an integer greater than 1, so the minimum value of x can be 2.

Let's assume x = 2, so 2 + (any prime number larger than 2) = 2 + odd = odd. Hence, x = 2 satisfies the criteria. However, x can be any even integer.

Say
x = 4 → 4 + (any prime number larger than 4) = 4 + odd = odd

As we do not have a unique value for x, the statement alone is not sufficient.

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