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If n is a positive integer, how many different factors n has? [#permalink]
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03 Jul 2017, 08:10
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If n is a positive integer, how many different factors n has? 1) N/5 is a prime number. 2) N has only two different prime factors.
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If n is a positive integer, how many different factors n has? [#permalink]
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03 Jul 2017, 10:13
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haardiksharma wrote: If n is a positive integer, how many different factors n has?
1) N/5 is a prime number. 2) N has only two different prime factors. Hi, Why is the Statement A not sufficient on a standalone basis? Since N/5 is "a" prime number, N can be written as (5^1) * (prime number ^1) and hence N has 2x2 = 4 different factors Am I making a mistake here? Thanks, Novice



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Re: If n is a positive integer, how many different factors n has? [#permalink]
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03 Jul 2017, 10:16



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Re: If n is a positive integer, how many different factors n has? [#permalink]
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22 Aug 2017, 10:36
hi ,
can anyone please explain second option here.
Howcome any interget has only two different prime factor.. Is this even correct?
Any number by default will have unity and number itself prime factor.. how "only two prime factor" makes sense here?
Now if we go ahead by assuming that, question is talking about prime factors..
then in that case we'll have four factor..wont we?



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Re: If n is a positive integer, how many different factors n has? [#permalink]
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25 Aug 2017, 19:54
1 n/5 = prime => n = prime x 5 all of the factor of n can be: 1, prime, 5, and n. So 4. But then if this prime number is 5, then n has only 3 factor: 1, 5, and n. INSUFF (also we can test 10 and 25 for this case)
2 N has only two different prime factors.
I understand this bolded phrase as: if n has 2 different prime factor 2 and 5, n can be written as: \(2^x5^y\) where total number of factor of n = (x+1)(y+1) Again INSUFF, because: x=1, y=1, n = 10: 4 factors x=2, y=1, n = 20: 6 factors
1+2: n = prime x5 n has only 2 different prime factors Combine together we can be sure the exponent of 5 and the other prime factor is 1, and that prime factor cant be 5. So n has 4 different factor. SUFF



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If n is a positive integer, how many different factors n has? [#permalink]
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25 Aug 2017, 20:24
haardiksharma wrote: If n is a positive integer, how many different factors n has?
1) N/5 is a prime number. 2) N has only two different prime factors. \(Statement 1:\) \(\frac{N}{5} = P\), or \(N = 5*P\) Now \(5\) is prime and if the other the prime no \(P = 5\), then \(N = 5^2\), hence we will have \((2+1) = 3\) factors: 1, 5 & 25 But if the other prime \(P = 3\), then \(N = 5*3\), hence we will have \((1+1)*(1+1) = 4\) factors: 1, 3, 5, 15 Hence \(Insufficient\) \(Statement 2:\) let's assume two different prime factors are \(2 & 3\). but we are not given what are the exponents of these prime numbers. for e.g if \(N = 2^1*3^1\), then the number of factors will be \((1+1)*(1+1) = 4\) (i.e 1, 2, 3 & 6), BUTif \(N = 2^2*3^1\), then the number of factors will be \((2+1)*(1+1)= 6\) (i.e 1, 2, 3, 4, 6 & 12) Hence \(Insufficient\) Combining 1 & 2 we know that \(N = 5*P\) and from statement 2 we know that \(N\) has two different prime factors, hence \(P\) is not equal to \(5\). so the number of factors will be \((1+1)*(1+1) = 4\) \(Sufficient\) Option \(C\)
Last edited by niks18 on 25 Aug 2017, 20:35, edited 1 time in total.



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If n is a positive integer, how many different factors n has? [#permalink]
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25 Aug 2017, 20:31
pratik1709 wrote: hi ,
can anyone please explain second option here.
Howcome any interget has only two different prime factor.. Is this even correct?
Any number by default will have unity and number itself prime factor.. how "only two prime factor" makes sense here?
Now if we go ahead by assuming that, question is talking about prime factors..
then in that case we'll have four factor..wont we? Hi pratik1709Statement two simply states that N has two different PRIME Factors and NOT two different factors which is very much possible for e.g \(6 = 2*3\), has only two prime factors \(2 & 3\), but has 4 factors = 1,2,3 & 6



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Re: If n is a positive integer, how many different factors n has? [#permalink]
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19 Sep 2017, 09:57
niks18 wrote: haardiksharma wrote: If n is a positive integer, how many different factors n has?
1) N/5 is a prime number. 2) N has only two different prime factors. \(Statement 1:\) \(\frac{N}{5} = P\), or \(N = 5*P\) Now \(5\) is prime and if the other the prime no \(P = 5\), then \(N = 5^2\), hence we will have \((2+1) = 3\) factors: 1, 5 & 25 But if the other prime \(P = 3\), then \(N = 5*3\), hence we will have \((1+1)*(1+1) = 4\) factors: 1, 3, 5, 15 Hence \(Insufficient\) \(Statement 2:\) let's assume two different prime factors are \(2 & 3\). but we are not given what are the exponents of these prime numbers. for e.g if \(N = 2^1*3^1\), then the number of factors will be \((1+1)*(1+1) = 4\) (i.e 1, 2, 3 & 6), BUTif \(N = 2^2*3^1\), then the number of factors will be \((2+1)*(1+1)= 6\) (i.e 1, 2, 3, 4, 6 & 12) Hence \(Insufficient\) Combining 1 & 2 we know that \(N = 5*P\) and from statement 2 we know that \(N\) has two different prime factors, hence \(P\) is not equal to \(5\). so the number of factors will be \((1+1)*(1+1) = 4\) \(Sufficient\) Option \(C\) hi Why it has to be assumed that, 5 and p are not raised to any power in the equation, N = 5 * P, when combining statement 1 and statement 2 together ....? thanks in advance ....



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Re: If n is a positive integer, how many different factors n has? [#permalink]
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19 Sep 2017, 10:37
gmatcracker2017 wrote: niks18 wrote: haardiksharma wrote: If n is a positive integer, how many different factors n has?
1) N/5 is a prime number. 2) N has only two different prime factors. \(Statement 1:\) \(\frac{N}{5} = P\), or \(N = 5*P\) Now \(5\) is prime and if the other the prime no \(P = 5\), then \(N = 5^2\), hence we will have \((2+1) = 3\) factors: 1, 5 & 25 But if the other prime \(P = 3\), then \(N = 5*3\), hence we will have \((1+1)*(1+1) = 4\) factors: 1, 3, 5, 15 Hence \(Insufficient\) \(Statement 2:\) let's assume two different prime factors are \(2 & 3\). but we are not given what are the exponents of these prime numbers. for e.g if \(N = 2^1*3^1\), then the number of factors will be \((1+1)*(1+1) = 4\) (i.e 1, 2, 3 & 6), BUTif \(N = 2^2*3^1\), then the number of factors will be \((2+1)*(1+1)= 6\) (i.e 1, 2, 3, 4, 6 & 12) Hence \(Insufficient\) Combining 1 & 2 we know that \(N = 5*P\) and from statement 2 we know that \(N\) has two different prime factors, hence \(P\) is not equal to \(5\). so the number of factors will be \((1+1)*(1+1) = 4\) \(Sufficient\) Option \(C\) hi Why it has to be assumed that, 5 and p are not raised to any power in the equation, N = 5 * P, when combining statement 1 and statement 2 together ....? thanks in advance .... Hi gmatcracker2017We know "\(P\)" is prime, hence it cannot be raised to any other power because in that case it will no more be a prime.. for eg if \(P=2\) and you raise it to power \(3\), then \(2^3 = 8\) = not prime




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