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If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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29 Jul 2016, 02:20
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If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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17 Jul 2017, 10:28
Official Solution:If \(n = x^5*y^7\), where \(x\) and \(y\) are positive integers greater than 1, then how many positive divisors does \(n\) have? (1) \(x\) does not have a factor \(p\) such that \(1 < p < x\) and \(y\) does not have a factor \(q\) such that \(1 < q < y\). This statement implies that both \(x\) and \(y\) are primes (a prime number does not have a factor which is greater than 1 and less than itself, it has only two factors 1 and itself). Now, if \(x\) and \(y\) are different primes, then the number of factors of \(n\) will be \((5+1)(7+1)=48\) but if \(x=y\), then \(n = x^5*y^7=x^{12}\) and it will have 13 factors. Not sufficient. (2) \(n\) has only two prime factors. If those primes are \(x\) and \(y\), then the number of factors of \(n\) will be \((5+1)(7+1)=48\) but if, say \(x=2*3=6\) and \(y=2\), then \(n = x^5*y^7=2^{12}*3^5\) and it will have \((12+1)(5+1)=78\) factors. Not sufficient. (1)+(2) Since from (2) \(n\) has only two prime factors, then from (1) it follows that \(x\) and \(y\) are different primes so the number of factors of \(n\) will be \((5+1)(7+1)=48\). Sufficient. Answer: C
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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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29 Jul 2016, 02:56
Bunuel wrote: If x and y are positive integers and \(n = x^5*y^7\), then how many positive divisors does n have?
(1) x does not have a factor p such that 1 < p < x and y does not have a factor q such that 1 < q < y. (2) n has only two prime factors. 1) implies both 'x' and 'y' are prime numbers , so n is expressed in its power of prime factors. Hence 1 is sufficient. 2) Although it says n has only two prime factors, they need not to be x and y, but also some root of x or y Hence not sufficient. My Answer 'A'



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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29 Jul 2016, 04:24
my take is D on this since first one talks on both X and Y being prime and so does B



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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29 Jul 2016, 08:49
If x and y are positive integers and \(n = x^5*y^7\), then how many positive divisors does n have?
the limiting factor here is x^5 as it will be paired with y^7 to find the number of positive divisors
(1) x does not have a factor p such that 1 < p < x and y does not have a factor q such that 1 < q < y. both x and y are prime here so the number of positive divisors will stay the same Sufficient
(2) n has only two prime factors. again, both x and y are prime here so the number of positive divisors will stay the same Sufficient
D



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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29 Jul 2016, 09:00
IMO E 1) x and y can be same also.its not given in question that x and y are diff positive integers
not suff
2)two diff primes if x=2 and y=3 then diff no of factors if x=6 and y=6 then diff no of factors
1+2,
same case as 2
so E



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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29 Jul 2016, 12:30
Bunuel wrote: If x and y are positive integers and \(n = x^5*y^7\), then how many positive divisors does n have?
(1) x does not have a factor p such that 1 < p < x and y does not have a factor q such that 1 < q < y. (2) n has only two prime factors. Statement 1: x does not have a factor p such that 1 < p < x and y does not have a factor q such that 1 < q < yMeans both x and y are prime Now , x could be 2 and y could be 3 in which case no of factors=6*8=48. But what if x=y, then no of factors=13.Insufficient Statement 2:(2) n has only two prime factorsMeaning x and y are different prime numbersSufficient Answer B



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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29 Jul 2016, 15:54
You can calculate the number of divisors of n, if you know that x and y are prime.
(1) x does not have a factor p such that 1 < p < x and y does not have a factor q such that 1 < q < y. The wording of this statement and the question stem implicitly states that x and y are different integers. Given this statement, x and y are distinct prime numbers, so the number of divisors of n can be calculated using this statement.
This statement is sufficient.
(2) n has only two prime factors. From the question stem \(n =x^5∗y^7\). This statement cannot be used to ascertain that x and y are the 2 prime factors of n. Let me illustrate with an example 
When \(n = 4^5 * 6^7\) and you are told that n has 2 prime factors, this does not mean 4 and 6 are the two prime factors. The 2 prime factors of n in this case are 2 and 3.
Hence, Statement 2 by itself is not sufficient to answer the question.
My answer is A.



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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31 Jul 2016, 22:33
Bunuel wrote: If x and y are positive integers and \(n = x^5*y^7\), then how many positive divisors does n have?
(1) x does not have a factor p such that 1 < p < x and y does not have a factor q such that 1 < q < y. (2) n has only two prime factors. very simple quetion. A says that both x and y are prime numbers. B mentions it clearly that x and y are prime numbers. so the answer will b D



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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31 Jul 2016, 22:35



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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01 Aug 2016, 00:08
The answer cannot be anything but D. First statement clearly means that both X & Y are prime . They can be equal as well as two separate prime numbers. If equal then no of factors would be 13 , if different then it would be 48. So insufficient . The second statement says that n has only 2 prime factors so X can be 2,4,8 and Y can be 3,9,27 ..all yielding different factors for n. Insufficient . But if we combine we surely know X & Y are prime and since n has only 2 prime factors both X & Y are different . Sufficient . Sent from my iPhone using GMAT Club Forum mobile app



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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01 Aug 2016, 00:14



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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01 Aug 2016, 12:06
Sorry it's a typo ..the correct answer has to be C (100%) Sent from my iPhone using GMAT Club Forum mobile app



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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01 Aug 2016, 23:06
Bunuel wrote: If x and y are positive integers and \(n = x^5*y^7\), then how many positive divisors does n have?
(1) x does not have a factor p such that 1 < p < x and y does not have a factor q such that 1 < q < y. (2) n has only two prime factors. The Question Asks for How many positive Divisors Does N Have. From Question Stem N = X ^ 5 * Y ^ 7 , X and Y are Positive Integers. Analyzing Statement 1 x does not have a factor p such that 1 < p < x and y does not have a factor q such that 1 < q < y.. Which say X and Y Are prime Numbers Lets Assume X = 2 Y= 5 then N = 2 ^ 5 * 5 ^ 7 which Equals to 32 * 78125 . Now this will have many Positive divisors. and No of Positive divisors will change with change of assumed prime Nos There 1 Is not Sufficient Analyzing Statement 2 n has only two prime factors Which means N = X ^5 * Y ^ 7 = P1 * P2 ( P1 and P 2 are Prime Nos ) But we do not know the Powers of prime Nos. For Example P1 =2 and P2 = 5 if the power of both prime no is one than N= 2 * 5 = 10 ( 10 Will have two divisors ) BUT If The Power of both prime nos is 2 Than N= 4 * 25 = 100 ( Now 100 will have more than 2 Divisors ) Therefore statement 2 is In Sufficient Analyzing Statement 1 And Statement 2 since we do not know the power of prime Nos. I will Go with option E Please correct if i am Wrong Thank You



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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06 Aug 2016, 00:47
Bunuel wrote: If \(x\) and \(y\) are positive integers and \(n = x^5\times y^7\), then how many positive divisors does n have? Bunuel, may I have some assistance with where I'm going wrong with (1+2)? (1) \(x\) does not have a factor \(p\) such that \(1 < p < x\) and \(y\) does not have a factor \(q\) such that \(1 < q < y\). \(x\) and \(y\) are either prime or 1. Assume both are prime for a counterexample. \(x = y = 2 \implies n = 2^{12} \text{ (}12 + 1 \text{ divisors})\\ x = 2, y = 3 \implies n = 2^53^7 ((5+1)(7+1) = 48 \text{ divisors})\) Insufficient(2)\(n\) has only two prime factors. Assuming this means two distinct prime factors, due to the integer constraint on \(x\) and \(y\), I do not believe that two prime factors is possible (setting x and y to 1 would fail, as would 1 and any other integer). \(x = 2, y = 3 \implies n = 2^53^7 ((5+1)(7+1) = 48 \text{ divisors})\\ x = 6, y = 2 \implies n = 3^52^{12} ((5+1)(12+1) = 72 \text{ divisors})\) Insufficient(1 + 2) (I can't for the life of me determine why this is not sufficient, none of the above answers seem correct to me). x and y must be positive integers (question stem) x or y may not be 1. (1) requires x and y are prime or 1, (2) requires that there are 2 prime factors, so we need two primes. x and y must therefore be prime (1) x and y must be distinct primes (2) \(x^k \times y^j\) will always have \((k+1)(j+1)\) divisors when \(x\) and \(y\) are distinct primes. Therefore sufficient.
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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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07 Aug 2016, 13:43
Argo wrote: DAllison2016 wrote: Bunuel wrote: If \(x\) and \(y\) are positive integers and \(n = x^5\times y^7\), then how many positive divisors does n have? Bunuel, may I have some assistance with where I'm going wrong with (1+2)? (1) \(x\) does not have a factor \(p\) such that \(1 < p < x\) and \(y\) does not have a factor \(q\) such that \(1 < q < y\). \(x\) and \(y\) are either prime or 1. Assume both are prime for a counterexample. \(x = y = 2 \implies n = 2^{12} \text{ (}12 + 1 \text{ divisors})\\ x = 2, y = 3 \implies n = 2^53^7 ((5+1)(7+1) = 48 \text{ divisors})\) Insufficient(2)\(n\) has only two prime factors. Assuming this means two distinct prime factors, due to the integer constraint on \(x\) and \(y\), I do not believe that two prime factors is possible (setting x and y to 1 would fail, as would 1 and any other integer). \(x = 2, y = 3 \implies n = 2^53^7 ((5+1)(7+1) = 48 \text{ divisors})\\ x = 6, y = 2 \implies n = 3^52^{12} ((5+1)(12+1) = 72 \text{ divisors})\) Insufficient(1 + 2) (I can't for the life of me determine why this is not sufficient, none of the above answers seem correct to me). x and y must be positive integers (question stem) x or y may not be 1. (1) requires x and y are prime or 1, (2) requires that there are 2 prime factors, so we need two primes. x and y must therefore be prime (1) x and y must be distinct primes (2) \(x^k \times y^j\) will always have \((k+1)(j+1)\) divisors when \(x\) and \(y\) are distinct primes. Therefore sufficient. I believe the answer is 'C'. Can someone explain why is this E? The answer is C. Sorry, E was a marked there by mistake.
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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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07 Aug 2016, 20:54
1<p<x: there is no p,get it: but doesnt it implies thAt x>1 @bunnel Sent from my Moto G (4) using GMAT Club Forum mobile app



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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08 Aug 2016, 04:11
Statement 1 is sufficient alone. If we know that X and Y are primes then obviously we can determine the number of factors of n. Total factors are (5+1)*(7+1)=48. But unfortunately the OA is C. Where is my mistake? Can you, bunuel, help me to understand the fact, please.



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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08 Aug 2016, 04:55
n = x^5 * y^7
Statement 1 means x and y are prime numbers. So if n can be written as n = (prime number)^5 * (another prime number)^7 then number of factors (divisors) of n would be (1+5)(1+7) = 48. Sufficient
Statement 2 says that n has exactly 2 prime factors but that doesn't necessitate that x and y be prime. Insufficient.
Why is the answer not A? Is it because we don't know whether x=y? Cos if it is so then n = x^5 * y^7 = x^12. In this case the number of factors will change to (1+12) = 13.



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Re: If n = x^5*y^7, where x and y are positive integers greater than 1 [#permalink]
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08 Aug 2016, 05:32
I'll give it a try. A) X, Y = prime. If X =/= Y then we would know the answer (6*8 factors). However, if X=Y then we would receive a lot of answers that would generate the same factors with the previous method. e.g. x=y=3 => 3^3*3^1=3^1*3^3. Thus, A is insufficient. As for B) (Which, in all honesty, was my first choice before I analyzed the question further). We now know that x and y consist of 2 prime numbers. IF x and y are two different prime numbers we would be able to answer the question. However(and here was my mistake), that information is given in (A). (In that case it would indeed be 6*8 factors). X and Y can still be different composite numbers without giving n more than 2 prime factors. e.g. x = 2*3 and y = 2*2*3. X and Y can also be composite numbers where x and y are two different prime numbers taken to any power. Thus, (as written above) We needed the information about x and y being prime numbers from (A) and we needed the information that n only had two different prime factors from (B). Thus > C is the answer. Great question, thanks!
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