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If x represents the number of positive factors of integer y, is x odd
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02 Oct 2014, 20:39
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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
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Re: If x represents the number of positive factors of integer y, is x odd
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03 Oct 2014, 00:59
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.
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Re: If x represents the number of positive factors of integer y, is x odd
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02 Oct 2014, 20:42
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.
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Re: If x represents the number of positive factors of integer y, is x odd
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02 Oct 2014, 23:48
aadikamagic wrote: If x represents the number of positive factors of integer y, is x odd?
y=n! where n is a positive integer greater than 1
y=m^2−1 where m is a positive integer greater than 1
a) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked b) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked c) Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient d) EACH statement ALONE is sufficient to answer the question asked e) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed 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
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Re: If x represents the number of positive factors of integer y, is x odd
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03 Oct 2014, 12:07
Bunuel wrote: 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. That's a nice explaination .



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Re: If x represents the number of positive factors of integer y, is x odd
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03 Oct 2014, 12:24
Bunuel wrote: 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. Thanks bunuel but could you please tell me what was the issue with my approach so that ill be careful on exam day
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If x represents the number of positive factors of integer y, is x odd
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03 Oct 2014, 12:26
aadikamagic wrote: 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=1m=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. (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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Re: If x represents the number of positive factors of integer y, is x odd
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03 Oct 2014, 12:27
Bunuel wrote: aadikamagic wrote: 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=1m=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. (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.
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Re: If x represents the number of positive factors of integer y, is x odd
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30 Aug 2015, 07:16
Bunuel wrote: 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. 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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If x represents the number of positive factors of integer y, is x odd
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30 Aug 2015, 07:29
goidelg wrote: Bunuel wrote: 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. 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. 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.



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