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How many prime factors does x^37 have?
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Updated on: 09 Mar 2015, 05:47
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37% (02:08) correct 63% (01:54) wrong based on 165 sessions
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How many prime factors does \(x^{37}\) have? (1) The number of prime factors of 16x is one more than the number of prime factors of \(x^4\) (2) \(2x^{16}\) has 2 prime factors
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Originally posted by TARGET730 on 09 Mar 2015, 05:38.
Last edited by Bunuel on 09 Mar 2015, 05:47, edited 1 time in total.
Edited the question.



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Re: How many prime factors does x^37 have?
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09 Mar 2015, 08:19
TARGET730 wrote: How many prime factors does \(x^{37}\) have?
(1) The number of prime factors of 16x is one more than the number of prime factors of \(x^4\)
(2) \(2x^{16}\) has 2 prime factors ans C.... 1) statement 1 tells us that 16x has one more prime factor than \(x^4\), which will be same as for x.... therefore it tells us that x is not multiple of 2... insufficient... 2) statement 2 tells us that \(2x^{16}\) has two prime factor .... it tells us that that one prime factor is 2 along with one more prime factor... does not tell us that x also has a 2 in it... insufficient... combined it tells us that x has only one prime factor that is itself.... sufficient
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Re: How many prime factors does x^37 have?
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09 Mar 2015, 13:31
Hi chetan2u, Can you please explain "combined it tells us that x has only one prime factor that is itself." ?
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Re: How many prime factors does x^37 have?
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09 Mar 2015, 13:52
How many prime factors does \(x^{37}\) have?Real GMAT question would mention that x is a positive integer. ALL GMAT divisiblity questions are limited to positive integers. So, I'd assume that it's given that x is a positive integer. First of all note that exponentiation does not "produce" primes, meaning that for positive integers m and n, m^n would have as many prime factors as m itself does. (1) The number of prime factors of 16x is one more than the number of prime factors of x^4 > 16x = 2^4*x. So it's given that 2^4*x has one more prime than x^4, which means that 2 is NOT a prime factor of x. We know nothing else about x. Not sufficient. (2) \(2x^{16}\) has 2 prime factors > this implies that x itself can have 1 prime (something other than 2) or 2 primes (2 and something other than 2). Not sufficient. (1)+(2) Since from (1) we know that 2 is NOT a factor of x, then from (2) it follows that x must have only 1 prime. Thus x^37 also will have only 1 prime. Sufficient. Answer: C. Hope it's clear.
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Re: How many prime factors does x^37 have?
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09 Mar 2015, 19:36
TARGET730 wrote: Hi chetan2u,
Can you please explain "combined it tells us that x has only one prime factor that is itself." ? hi TARGET7301) statement 1 tells us that 16x has one more prime factor than x^4, which will be same as for x.... therefore it tells us that x is not multiple of 2... insufficient... the power will have same prime factors as x itself2) statement 2 tells us that 2x^{16} has two prime factor .... it tells us that that one prime factor is 2 along with one more prime factor... does not tell us that x also has a 2 in it... insufficient... this tells us that 2x^{16} has two factors.. one is two and other could be anything.... but x may be multiple of two then x will have 2 prime factors... if x is not multiple of 2 then x will have only one prime factor...combined it tells us that x has only one prime factor that is itself.... sufficient now 2 is not in x so x contains one prime factor which can be itself or some root of x...
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How many prime factors does x^37 have?
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Updated on: 12 Mar 2015, 20:31
Here's my explanation, I tried to keep it simple:
The question is essentially asking us if x is a prime, or can be factored into a single prime.
St (1): tells us that 2 is not a factor of x. We know this because if 16(x) has one less prime factor than x, that "one less prime factor" is 2. However, we still do not know whether x is prime or not. N.S. 2^8 =16, for those who wonder how that association is made.
For example:
x can equal 15 which satisfies the condition in st(1), but doesn't solve the problem.
x can also equal 3 which satisfies the condition in st(1) and DOES solve the problem.
St (2) tells us that 2(x)^16 only has two primes. This means that x can be factored into a single prime number OR x has 2 and another prime as its prime factors.
St(1) and St (2) tell us that, working with St(2) first; 2(x)^16 only has two primes. St(1) tells us that 2 is not a factor of x. Therefore x must be a prime number must be a prime number or an integer that can be factored into a single prime number, other than 2.
C.
Posted from my mobile device
Originally posted by ak1802 on 10 Mar 2015, 09:04.
Last edited by ak1802 on 12 Mar 2015, 20:31, edited 1 time in total.



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Re: How many prime factors does x^37 have?
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10 Mar 2015, 09:10
ak1802 wrote: Here's my explanation, I tried to keep it simple:
The question is essentially asking us if x is a prime, or can be factored into a single prime.
St (1): tells us that 2 is not a factor of x. We know this because if 16(x) has one less prime factor than x, that "one less prime factor" is 2. However, we still do not know whether x is prime or not. N.S. 2^8 =16, for those who wonder how that association is made.
For example:
x can equal 15 which satisfies the condition in st(1), but doesn't solve the problem.
x can also equal 3 which satisfies the condition in st(1) and DOES solve the problem.
St (2) tells us that 2(x)^16 only has two primes. This means that x can be factored into a single prime number OR x has 2 and another prime as its prime factors.
St(1) and St (2) tell us that, working with St(2) first; 2(x)^16 only has two primes. St(1) tells us that 2 is not a factor of x. Therefore x must be a prime number.
C.
Posted from my mobile device hi, it is not necessary x will be a prime number ... it can be a square of a prime number... only surety is that it has only one prime number as factor
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How many prime factors does x^37 have?
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10 Mar 2015, 10:24
chetan2u wrote: ak1802 wrote: Here's my explanation, I tried to keep it simple:
The question is essentially asking us if x is a prime, or can be factored into a single prime.
St (1): tells us that 2 is not a factor of x. We know this because if 16(x) has one less prime factor than x, that "one less prime factor" is 2. However, we still do not know whether x is prime or not. N.S. 2^8 =16, for those who wonder how that association is made.
For example:
x can equal 15 which satisfies the condition in st(1), but doesn't solve the problem.
x can also equal 3 which satisfies the condition in st(1) and DOES solve the problem.
St (2) tells us that 2(x)^16 only has two primes. This means that x can be factored into a single prime number OR x has 2 and another prime as its prime factors.
St(1) and St (2) tell us that, working with St(2) first; 2(x)^16 only has two primes. St(1) tells us that 2 is not a factor of x. Therefore x must be a prime number.
C.
Posted from my mobile device hi, it is not necessary x will be a prime number ... it can be a square of a prime number... only surety is that it has only one prime number as factor Yes, you're correct. I stated above that the question is asking if X can either be prime, or factored into a single prime. [edit] didn't notice the red text on my phone. You're right, error on my part in St(1) and St(2). X must be an integer, that can be factored into a single prime number, other than 2. Posted from my mobile device



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Re: How many prime factors does x^37 have?
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05 Dec 2016, 22:40
Such an Amazing Question. Here is my Approach >
Wee need the number of prime factors of \(X^{37}\) RULE > Raising a positive integer to a positive exponent does not change its prime factors. Hence \(X^{37}\) will have the exact same prime factors as \(X\). Thus in a way the Question is asking for the => number of prime factors of \(X\).
Lets Dive into statements
Statement 1> \(16X\)=> one more prime factor than \(X^4\) Again X^4 will have the exact same prime factors as \(X\). So \(16X\) has one more prime than X \(16X=2^4*X\) => That additional Prime must be 2. Thus \(X\) does have 2 as its Prime factor. But we don't know the number of Prime factors of X. Hence Not sufficient.
Statement 2> \(2X^{16}\) has two Prime factors. This means that \(X\) can have either one prime factor (exclusive of 2) Or it can have Two prime factors(inclusive of 2) Hence not sufficient.
Combing the two statements> We can say that X does have two as its prime=>\(2*X^{16}\) has two primes => \(X\) must have only one prime. Hence Sufficient
Hence C
Great Question..!!!
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Re: How many prime factors does x^37 have?
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Re: How many prime factors does x^37 have?
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