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# M31-13

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Math Expert
Joined: 02 Sep 2009
Posts: 49968

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09 Jun 2015, 07:59
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Difficulty:

75% (hard)

Question Stats:

58% (01:18) correct 42% (01:36) wrong based on 74 sessions

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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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Joined: 02 Sep 2009
Posts: 49968

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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.

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Joined: 03 Jul 2014
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07 Oct 2016, 01:43
1
------, 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
Intern
Joined: 13 Nov 2016
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07 Dec 2016, 04:01
1
Had exactly the same issue as above.
Intern
Joined: 29 Jul 2016
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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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Joined: 26 Dec 2011
Posts: 195
Location: United States (NY)
Concentration: Finance, Entrepreneurship
GPA: 3.4
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12 Oct 2017, 05:01
1
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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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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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.
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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19 Nov 2017, 02:46
1
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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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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07 May 2018, 03:41
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 08 Sep 2016
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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?
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Schools: IIMA , IIMA PGPX"20
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31 Jul 2018, 14:14
where xx is a positive integer greater than 1, such that xx does NOT have a factor pp such that 1<p<x1<p<x

Bunuel,
How do we infer that X is a prime number, can you please elaborate.
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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31 Jul 2018, 20:27
gsingh0711 wrote:
where xx is a positive integer greater than 1, such that xx does NOT have a factor pp such that 1<p<x1<p<x

Bunuel,
How do we infer that X is a prime number, can you please elaborate.

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Re: M31-13 &nbs [#permalink] 31 Jul 2018, 20:27
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# M31-13

Moderators: chetan2u, Bunuel

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