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(1): 3! < n < 4! → 6 < n < 24. Every even integer can be expressed as a multiple of 2, all even numbers in range will give us a "yes" to question. However, since primes are in range (7, 11, 13, 17, 19, 23), that cannot be factored further than itself, we can also answer question "no." If n = 8, it has 2 integer factors 2 and 4 that fit criteria: 1 < 2 < 4 < 8 → "yes" If n = 7, it cannot be broken down into factors that fit criteria of question → "no" Insufficient (2): Any positive, odd multiple of 3 is a possible value for n. For larger multiples of 3, we can easily fit criteria, giving "yes" answer. However, since any number’s smallest multiple is itself, 3 is a possible value for n that would not fit criteria, giving "no." If n = 15, we can factor it as 3 and 5, and we fit criteria: 1< 3 < 5 < 15 → "yes" If n = 9, we can only factor it as 3 and 3, which does not fit criteria → "no" Insufficient Combined: (1) limits possible values of n to those integers between 6 and 24. (2) adds limitation that n be odd multiple of 3. Possible values for n = (9, 15 and 21) If n = 9, we answer question "no," If n = 15 or 21, we can answer question "yes." Insufficient

Hi, I want to know if we have more simple solution for this question, please.

(1): 3! < n < 4! → 6 < n < 24. Every even integer can be expressed as a multiple of 2, all even numbers in range will give us a "yes" to question. However, since primes are in range (7, 11, 13, 17, 19, 23), that cannot be factored further than itself, we can also answer question "no." If n = 8, it has 2 integer factors 2 and 4 that fit criteria: 1 < 2 < 4 < 8 → "yes" If n = 7, it cannot be broken down into factors that fit criteria of question → "no" Insufficient (2): Any positive, odd multiple of 3 is a possible value for n. For larger multiples of 3, we can easily fit criteria, giving "yes" answer. However, since any number’s smallest multiple is itself, 3 is a possible value for n that would not fit criteria, giving "no." If n = 15, we can factor it as 3 and 5, and we fit criteria: 1< 3 < 5 < 15 → "yes" If n = 9, we can only factor it as 3 and 3, which does not fit criteria → "no" Insufficient Combined: (1) limits possible values of n to those integers between 6 and 24. (2) adds limitation that n be odd multiple of 3. Possible values for n = (9, 15 and 21) If n = 9, we answer question "no," If n = 15 or 21, we can answer question "yes." Insufficient

Hi, I want to know if we have more simple solution for this question, please.

Does the integer n have two factors, x and y, such that 1 < x < y < n?

(1) 3! < n < 4! --> 6 < n < 24 --> If n=7=prime, then the answer is NO but if n=10, then the answer is YES. Not sufficient.

(2) n is odd and a multiple of 3. If n=3=prime, then the answer is NO but if n=15, then the answer is YES. Not sufficient.

(1)+(2) The same here: n=9=3^2 gives a NO answer, while n=15=3*5 gives an YES answer. Not sufficient.

(1): 3! < n < 4! → 6 < n < 24. Every even integer can be expressed as a multiple of 2, all even numbers in range will give us a "yes" to question. However, since primes are in range (7, 11, 13, 17, 19, 23), that cannot be factored further than itself, we can also answer question "no." If n = 8, it has 2 integer factors 2 and 4 that fit criteria: 1 < 2 < 4 < 8 → "yes" If n = 7, it cannot be broken down into factors that fit criteria of question → "no" Insufficient (2): Any positive, odd multiple of 3 is a possible value for n. For larger multiples of 3, we can easily fit criteria, giving "yes" answer. However, since any number’s smallest multiple is itself, 3 is a possible value for n that would not fit criteria, giving "no." If n = 15, we can factor it as 3 and 5, and we fit criteria: 1< 3 < 5 < 15 → "yes" If n = 9, we can only factor it as 3 and 3, which does not fit criteria → "no" Insufficient Combined: (1) limits possible values of n to those integers between 6 and 24. (2) adds limitation that n be odd multiple of 3. Possible values for n = (9, 15 and 21) If n = 9, we answer question "no," If n = 15 or 21, we can answer question "yes." Insufficient

Hi, I want to know if we have more simple solution for this question, please.

Does the integer n have two factors, x and y, such that 1 < x < y < n?

(1) 3! < n < 4! --> 6 < n < 24 --> If n=7=prime, then the answer is NO but if n=10, then the answer is YES. Not sufficient.

(2) n is odd and a multiple of 3. If n=3=prime, then the answer is NO but if n=6, then the answer is YES. Not sufficient.

(1)+(2) The same here: n=9=3^2 gives a NO answer, while n=15=3*5 gives an YES answer. Not sufficient.

Does the integer n have two factors, x and y, such that 1 < x < y < n?

(1) 3! < n < 4! (2) n is odd and a multiple of 3.

The question is not difficult if you understand the theory of factors properly.

Does n have two factors x and y such that x and y lie between 1 and n and are distinct? When does a number have factors between 1 and itself? When it is a composite (not a prime) number. Every composite number has a factor in between 1 and itself. When will the factors be distinct i.e. when does the number have more than 1 factors? When it is not a perfect square or a prime number. A perfect square of a prime number such as 4 has only 1 factor between 1 and itself (1, 2, 4).

So we want two things in our n : It should not be prime and it should not be square of a prime number.

(1) 3! < n < 4! This means 6 < n < 24 If n is 7, it is prime. It has no x and y. If it is 8 it is not a prime and not a square of a prime. It has x and y. Not sufficient

(2) n is odd and a multiple of 3. If n is 3, it is prime. It has no x and y. If it is 15, it is not a prime and not a square of a prime. It has x and y. Not sufficient

Using both, n could be 9/12/15 etc 9 is the square of a prime. It has no x and y. 12 is not a prime and not the square of a prime. It has x and y. Not sufficient.

Re: Does integer n have 2 factors x & y such that 1 < x < y < n? [#permalink]

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15 Mar 2016, 09:36

Here we are asked whether n has more than two factors excluding 1 and n such that 1 < x < y < n now statement 1 => n=> 96,24) for 7=> NO for 20=> YES hence not sufficient Statement 2 => N=3 => NO N= 9=> No N= 27 => YES hence Not sufficient Combining them N=> 9=> NO N= 18 => YES hence E
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Could you please clarify the highlighted part below about statement 2? I must not be understanding what the condition "n is odd" implies. I thought it could only be 3-9-15-21, but not 6-12-18, etc.

Thanks!

(2) n is odd and a multiple of 3. If n is 3, it is prime. It has no x and y. If it is 6 it is not a prime and not a square of a prime. It has x and y. Not sufficient

VeritasPrepKarishma wrote:

goodyear2013 wrote:

Does the integer n have two factors, x and y, such that 1 < x < y < n?

(1) 3! < n < 4! (2) n is odd and a multiple of 3.

The question is not difficult if you understand the theory of factors properly.

Does n have two factors x and y such that x and y lie between 1 and n and are distinct? When does a number have factors between 1 and itself? When it is a composite (not a prime) number. Every composite number has a factor in between 1 and itself. When will the factors be distinct i.e. when does the number have more than 1 factors? When it is not a perfect square or a prime number. A perfect square of a prime number such as 4 has only 1 factor between 1 and itself (1, 2, 4).

So we want two things in our n : It should not be prime and it should not be square of a prime number.

(1) 3! < n < 4! This means 6 < n < 24 If n is 7, it is prime. It has no x and y. If it is 8 it is not a prime and not a square of a prime. It has x and y. Not sufficient

(2) n is odd and a multiple of 3. If n is 3, it is prime. It has no x and y. If it is 6 it is not a prime and not a square of a prime. It has x and y. Not sufficient

Using both, n could be 9/12/15 etc 9 is the square of a prime. It has no x and y. 12 is not a prime and not the square of a prime. It has x and y. Not sufficient.

Could you please clarify the highlighted part below about statement 2? I must not be understanding what the condition "n is odd" implies. I thought it could only be 3-9-15-21, but not 6-12-18, etc.

Thanks!

(2) n is odd and a multiple of 3. If n is 3, it is prime. It has no x and y. If it is 6 it is not a prime and not a square of a prime. It has x and y. Not sufficient

VeritasPrepKarishma wrote:

goodyear2013 wrote:

Does the integer n have two factors, x and y, such that 1 < x < y < n?

(1) 3! < n < 4! (2) n is odd and a multiple of 3.

The question is not difficult if you understand the theory of factors properly.

Does n have two factors x and y such that x and y lie between 1 and n and are distinct? When does a number have factors between 1 and itself? When it is a composite (not a prime) number. Every composite number has a factor in between 1 and itself. When will the factors be distinct i.e. when does the number have more than 1 factors? When it is not a perfect square or a prime number. A perfect square of a prime number such as 4 has only 1 factor between 1 and itself (1, 2, 4).

So we want two things in our n : It should not be prime and it should not be square of a prime number.

(1) 3! < n < 4! This means 6 < n < 24 If n is 7, it is prime. It has no x and y. If it is 8 it is not a prime and not a square of a prime. It has x and y. Not sufficient

(2) n is odd and a multiple of 3. If n is 3, it is prime. It has no x and y. If it is 6 it is not a prime and not a square of a prime. It has x and y. Not sufficient

Using both, n could be 9/12/15 etc 9 is the square of a prime. It has no x and y. 12 is not a prime and not the square of a prime. It has x and y. Not sufficient.

Answer (E)

Yes, you are right. Consider only the odd multiples. n = 15 will have x and y. Edited.
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