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a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer

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Re: a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer  [#permalink]

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06 Apr 2017, 00:38
Bunuel wrote:
Bunuel wrote:
a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer such that it 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

Kudos for a correct solution.

OFFICIAL SOLUTION:

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 a/x to be an integer where x is a prime, x can take 3 values: 2, 5, or 31.

Hi Bunnel,

Can you please clarify if the value of x = 2, then according to 1 < p < x, what will be the value of p. I marked option C because x has to be a prime number and I assumed p = 2 as its given 1 < p < x and x has to be greater than p
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Joined: 02 Sep 2009
Posts: 58310
Re: a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer  [#permalink]

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06 Apr 2017, 00:55
sujay840 wrote:
Bunuel wrote:
Bunuel wrote:
a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer such that it 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

Kudos for a correct solution.

OFFICIAL SOLUTION:

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 a/x to be an integer where x is a prime, x can take 3 values: 2, 5, or 31.

Hi Bunnel,

Can you please clarify if the value of x = 2, then according to 1 < p < x, what will be the value of p. I marked option C because x has to be a prime number and I assumed p = 2 as its given 1 < p < x and x has to be greater than p

We are told that x is a positive integer such that it does NOT have a factor p such that 1 < p < x.

2 does not have a factor p such that 1 < p < 2.
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Re: a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer  [#permalink]

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02 Mar 2018, 10:44
Bunuel wrote:
a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer greater than 1, such that it 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

We can start by simplifying a:

5^15 - 625^3 = 5^15 - (5^4)^3 = 5^15 - 5^12 = 5^12(5^3 - 1) = 5^12(124) = 5^12(4)(31) = 5^12( 2^2)(31)

If a/x is an integer, then x is a factor of a. However, if x does not have a factor p such that 1 < p < x, then x must be prime number. For example, if x = 5, we see that x doesn’t have a factor between 1 and itself. Since a has three distinct prime factors, there are three distinct values for x: 2, 5 and 31.

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Re: a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer  [#permalink]

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01 Jul 2019, 08:27
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer   [#permalink] 01 Jul 2019, 08:27

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