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aadikamagic
Joined: 03 Jul 2013
Last visit: 04 Dec 2017
Posts: 72
Schools: ISB '17 (A) IIM (A)
GMAT 1: 660 Q48 V32
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
56% (02:07) correct
44%
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wrong
based on 537
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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
(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
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
aadikamagic
Joined: 03 Jul 2013
Last visit: 04 Dec 2017
Posts: 72
Schools: ISB '17 (A) IIM (A)
GMAT 1: 660 Q48 V32
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.
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.
aadikamagic
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
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.
_________________
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.
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Posts: 3
Bunuel
That's a nice explaination .
aadikamagic
Joined: 03 Jul 2013
Last visit: 04 Dec 2017
Posts: 72
Schools: ISB '17 (A) IIM (A)
GMAT 1: 660 Q48 V32
Bunuel
Thanks bunuel but could you please tell me what was the issue with my approach so that ill be careful on exam day
aadikamagic
(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.
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aadikamagic
Joined: 03 Jul 2013
Last visit: 04 Dec 2017
Posts: 72
Schools: ISB '17 (A) IIM (A)
GMAT 1: 660 Q48 V32
Bunuel
Ohh god I feel so stupid now. Thanks a ton man.
Bunuel
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.
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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.
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.
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.
Posts: 136
aadikamagic
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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