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M31-13

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M31-13 [#permalink]

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New post 09 Jun 2015, 07:59
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\(a = 5^{15} - 625^3\) and \(\frac{a}{x}\) is an integer, where \(x\) is a positive integer greater than 1, such that \(x\) does NOT have a factor \(p\) such that \(1 < p < x\), then how many different values for \(x\) are possible?

A. None
B. One
C. Two
D. Three
E. Four

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Re M31-13 [#permalink]

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New post 09 Jun 2015, 08:00
Official Solution:

\(a = 5^{15} - 625^3\) and \(\frac{a}{x}\) is an integer, where \(x\) is a positive integer greater than 1, such that \(x\) does NOT have a factor \(p\) such that \(1 < p < x\), then how many different values for \(x\) are possible?

A. None
B. One
C. Two
D. Three
E. Four


First of all, notice that \(x\) is a positive integer such that it does NOT have a factor \(p\) such that \(1 < p < x\) simply means that \(x\) is a prime number.

Next, \(a = 5^{15} - 625^3=5^{15} - 5^{12}=5^{12}(5^3-1)=5^{12}*124=2^2*5^{12}*31\).

Finally, for \(\frac{a}{x}\) to be an integer where \(x\) is a prime, \(x\) can take 3 values: 2, 5, or 31.


Answer: D
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Re: M31-13 [#permalink]

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New post 07 Oct 2016, 01:43
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------, where xx is a positive integer greater than 1, such that it does NOT have a factor p-------

The it in the question is ambiguous. I took a/x which is an integer as not having a factor p.

The question will be much clear if the it is replaced with x
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Re: M31-13 [#permalink]

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New post 07 Dec 2016, 04:01
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Had exactly the same issue as above.
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Re: M31-13 [#permalink]

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New post 08 Jun 2017, 10:02
I think this question would be much clearer if the word 'any' was added on: "where x is a positive integer greater than 1, such that it does NOT have ANY factor p". Or else how would one interpret P as being associated to one value, which is a factor P. If it were to say ANY, then P can take a range of value, which makes it clear that X is a prime.
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Re: M31-13 [#permalink]

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New post 12 Oct 2017, 05:01
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saumya2810 wrote:
------, where xx is a positive integer greater than 1, such that it does NOT have a factor p-------

The it in the question is ambiguous. I took a/x which is an integer as not having a factor p.

The question will be much clear if the it is replaced with x


Agreed with this.
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Re: M31-13 [#permalink]

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New post 12 Oct 2017, 05:06
okay wrote:
saumya2810 wrote:
------, where xx is a positive integer greater than 1, such that it does NOT have a factor p-------

The it in the question is ambiguous. I took a/x which is an integer as not having a factor p.

The question will be much clear if the it is replaced with x


Agreed with this.


Edited as suggested. Thank you.
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Re: M31-13 [#permalink]

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New post 19 Nov 2017, 01:39
a factor pp such that 1<p<x 1<p<x simply means that xx is a prime number.
How x is a prime number.?

Can you please explain with example.
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Re: M31-13 [#permalink]

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New post 19 Nov 2017, 02:46
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abhinashgc wrote:
a factor pp such that 1<p<x 1<p<x simply means that xx is a prime number.
How x is a prime number.?

Can you please explain with example.


A prime number is a positive integer with exactly two factors: 1 and itself. If an integer does not have a factor which is greater than 1 and less than itself, then it must be a prime. For example, 5 (a prime number) does not have a factor which is more than 1 and less than 5 but 4 has such factor: 2.
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Re: M31-13 [#permalink]

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New post 19 Nov 2017, 22:09
Bunuel wrote:
abhinashgc wrote:
a factor pp such that 1<p<x 1<p<x simply means that xx is a prime number.
How x is a prime number.?

Can you please explain with example.


A prime number is a positive integer with exactly two factors: 1 and itself. If an integer does not have a factor which is greater than 1 and less than itself, then it must be a prime. For example, 5 (a prime number) does not have a factor which is more than 1 and less than 5 but 4 has such factor: 2.


Thanks a lot bb
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Re M31-13 [#permalink]

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New post 07 May 2018, 03:41
I think this is a high-quality question and I agree with explanation.
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Re: M31-13 [#permalink]

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New post 16 Jun 2018, 08:39
when x=31 , 1<p<31 => p=2,3.... and 31 is prime
when x=2 if 1<p<2 what is the value of p here?
Re: M31-13   [#permalink] 16 Jun 2018, 08:39
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