If x represents the number of positive factors of integer y, is x odd

My Aprproach:

Statement 1)

when n=2, y=2, factors of 2 are 1,2 so x=2
when n=3, y=6, factors of 6 are 1,2,3,6 so x=4
n=4,y=24, factors of y are 1,2,3,4,6,8,12,24 x= 8


hence x can never be odd. Sufficient.

Statement 2)

m=2, y=1, factors of y =1, so x=1
m=3, y=8, factors of y=1,2,4,8 x=4

Insufficient as x can be even or odd.

IMO A but it doesnt match the official answer. Bunuel please help.

Y = n! = Since n > 1, What ever be the number, the number of factors of n! will always be even. Here's how:

For any number n, n! = 1 * 2 * 3 * 4.... n

No. of factors of 1 = 1 i.e 1
No. of factors of 2 = 2 i.e 1, 2
No. of factors of 3 = 2 i.e 1, 3
etc. etc.

This would mean that what ever be the numbers ahead and what ever be their number of factors (Even / Odd), they would always be multiplied by an even number. Since the number of factors are calculated by multiplying the number of factors of each prime factor.

So, A will lead to a answer Yes.

Now the correct answer to this question will either be A or D. To analyze it, lets look at choice B.

y=m^2−1 m>1

y = (m + 1) (m - 1)
This too will always lead to an even choice since either of m + 1 and m - 1 will have even number of factors.

Therefore, this can be answered by using either of choices.

Ans. D

That's a nice explaination .

Thanks bunuel but could you please tell me what was the issue with my approach so that ill be careful on exam day

(2) y = m^2 − 1 where m is a positive integer greater than 1

If m = 2, then y = 2^2 - 1 = 3, not 1. Factors of 3 are 1 and 3.

Ohh god I feel so stupid now. Thanks a ton man.

Sorry guys, I just don't get it.
Take 36 for example - 36= 2^2*3^2.
Two prime, distinct factors.

It must be some weird terminology issue, I would be glad if someone could explain this to me (Taking the GMAT on wednsday!)

Thanks in advance.

1) y=n! where n is a positive integer greater than 1-

let put n=2 , factors =2 =x,
n=3 , factors (1+1)(1+1) = x = 4
hence always x = even ....sufficient - for any number n>1

2) y=m2−1 where m is a positive integer greater than 1

put m=2 , y =3 factors = 2 =x
m=3 , y= 8 factors = 4.
hence always x = even ---- sufficient for any value m>1

hence option D correct.

This is a very good question!

Logic used Number of factors of any number = product of the number of factors of its prime factors
Eg: 6 = 3*2;
Number of factors of 3 is 2 {1,3}
Number of factors of 2 is 2 {1,2}
Therefore, number of factors of 6 = 2*2 =4 {1,2,3,6}

Now evaluating the statements:

Statement 1: y = n! and n is an integer greater than 1
Number of factors of 2 is 2 (even)
Therefore, number of factors of n!, with n>1, = 2*(some integer) = even

Therefore, no matter what, the total number of factors of n! with n>1 is even

Statement 1 is sufficient


Statement 2: y = m^2 − 1 where m is a positive integer greater than 1
y=(m-1)*(m+1)
Please note that m is greater than 1
Therefore, m-1>0

Case 1: m=2
Therefore, y=(2-1)(2+1) = 1*3
Number of factors of 3 {1,3} is 2. Therefore, even

Case 2: m>2
Therefore, m-1>1
In this case, undoubtedly, y is a product of prime numbers (if y is a product of composite numbers, then composite numbers are a product of prime numbers)
Therefore, y=prime*(any integer)
Number of factors of y = (number of factors of a prime number)*(some integer)
= 2*(some integer)
= even
Therefore, no matter what, number of factors of y are ever i.e. x is even
Statement 2 is sufficient

Therefore, (D)

Mool Mantra: Only 1 and perfect even powers of prime numbers have odd factors. Rest all have even factors

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