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

Question

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

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

Bunuel · Accepted Answer

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

The question asks whether the number of factors of y is odd. The  number of distinct factors  of a perfect square is ALWAYS ODD.   The reverse is also true : if a number has the odd number of distinct factors then it's a perfect square . So, the question asks whether y is a perfect square.

(1) y = n! where n is a positive integer greater than 1. Among factorials, only 0! and 1! are perfect squares. So, y is not. Sufficient.

(2) y = m^2 − 1 where m is a positive integer greater than 1 --> y is 1 less, than a perfect square, so not a perfect square (two positive consecutive integers cannot both be prefect squares). Sufficient.

Answer: D.

ENGRTOMBA2018 · Answer

Look below:

For any given number, N, the number of factors of N are as follows:

N = a^p * b^q * c^r .... Where a,b,c are prime number (=2,3,5,7,11...) and p,q,r are integers.  This should always be the first step.

Writing N in terms of its prime factors is known as prime factorization.

Once you have done the prime factorization, the number of factors of N= (p+1)(q+1)(r+1)...  (Please note that the formula for number of factors include 1 and the number itself as well.)

What you are confusing is the definition of

1. Prime factors
2. Positive factors

You are correct that 36 has 2 prime factors but for positive factors, you need to use the formula mentioned above.

So once you write 36 =  2^2 * 3^2 , you need to take the powers of the prime factors (2 and 3 in this case) and calculate (power of 2 +1)(power of 3 +1) = (2+1)(2+1) = 3*3 = 9, an odd number.

By counting, the factors of 36 are:
1 36
2 18
3 12
4 9
6  6  (so in all you have 9 factors, with 6 only counted ONCE).

The above property of total factors to be odd are especially true for perfect squares. The reverse is true as well.

Thus, for a perfect square, the total number of positive factors will always be ODD while number with ODD number of positive factors will ALWAYS be perfect squares (25,36,49, etc)

Hope this helps.

aadikamagic · Answer

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.

Thoughtosphere · Answer

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 :done

pragati0007 · Answer

That's a nice explaination .

aadikamagic · Answer

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

Bunuel · Answer

(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.

aadikamagic · Answer

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

goidelg · Answer

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.

JiddiHuMain · Answer

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.

adityaganjoo · Answer

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

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

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