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# How many prime factors does x^37 have?

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Manager
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How many prime factors does x^37 have? [#permalink]

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Updated on: 09 Mar 2015, 05:47
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34% (01:35) correct 66% (07:04) wrong based on 154 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? [#permalink]

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09 Mar 2015, 08:19
1
1
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? [#permalink]

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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? [#permalink]

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09 Mar 2015, 13:52
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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.

Hope it's clear.
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Re: How many prime factors does x^37 have? [#permalink]

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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 TARGET730

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...
the power will have same prime factors as x itself
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...
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? [#permalink]

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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.

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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? [#permalink]

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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? [#permalink]

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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.

 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? [#permalink]

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05 Dec 2016, 22:40
1
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? [#permalink]

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Re: How many prime factors does x^37 have?   [#permalink] 15 Jan 2018, 09:25
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