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Does integer n have 2 factors x & y such that 1 < x < y < n?
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 13 Jan 2014, 03:57
Question Stats:
59% (01:48) correct 41% (01:45) wrong based on 357 sessions
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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. OE (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.
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Does integer n have 2 factors x & y such that 1 < x < y < n?
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13 Jan 2014, 05:31
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. OE (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 --> 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. Answer: E.
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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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13 Jan 2014, 05:32
Bunuel 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. OE (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 --> 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. Answer: E. Similar questions: does-the-integer-k-have-a-factor-p-such-that-1-p-k-126735.htmlif-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html
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Does integer n have 2 factors x & y such that 1 < x < y < n?
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17 Mar 2015, 21:21
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 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. Answer (E)
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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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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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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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02 Dec 2017, 14:09
Hi VeritasPrepKarishma, 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) |
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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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03 Dec 2017, 23:21
Hadrienlbb wrote: Hi VeritasPrepKarishma, 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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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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15 Mar 2019, 13:05
VeritasKarishma doesnt the question imply that are there two factors x,y so that 1<x<y<n. doesnt it mean that " are there any two such factors so that 1<x<y<n". Shouldnt the ans. be yes. Say ( x=2, y =5) . such value exist? why are we saying 1,7 wil statisfy as 1 is not less than x(x=1) in this case and it violates the condition mentioned in question. Please clarify. too confused. thanks
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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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16 Mar 2019, 06:56
Mudit27021988 wrote: VeritasKarishma doesnt the question imply that are there two factors x,y so that 1<x<y<n. doesnt it mean that " are there any two such factors so that 1<x<y<n". Shouldnt the ans. be yes. Say ( x=2, y =5) . such value exist? why are we saying 1,7 wil statisfy as 1 is not less than x(x=1) in this case and it violates the condition mentioned in question. Please clarify. too confused. thanks This is not given to you. It is asked. Question: Does the integer n have two factors, x and y, such that 1 < x < y < n? It does not give us that n has 2 factors. It asks us if it does. (1) 3! < n < 4! All you know here is that 6 < n < 24 We don't know that n has 2 such factors. We just know that n could be 7 or 8 or 9 etc. In some cases it will have such 2 factors, in others it won't.
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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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16 Mar 2019, 08:31
Question: Does the integer n have two factors, x and y, such that 1 < x < y < n?
Restatement: does n have minimum 4 factors?
I would start from statement 2
n may have following values.
3 (2 factors: no to main q)
9 (3 factors: no to main q)
15(4 factors: Yes to main q)
INSUFFICIENT
Statement 1: 6<n<24
9 & 15 (as already calculated) fall in between the mentioned range.
9 (3 factors: no to main q)
15 (4 factors: Yes to main q)
INSUFFICIENT
Combining:
9 & 15 all suffice statement 1&2.
9 (3 factors: no to main q)
15 (4 factors: Yes to main q)
INSUFFICIENT.
E is my answer.
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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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Re: Does integer n have 2 factors x & y such that 1 < x < y < n?
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01 Jun 2020, 05:31
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