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

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New post 15 May 2015, 03:50
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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

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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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New post 18 May 2015, 07:10
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5
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.

Answer: D.
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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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New post Updated on: 15 May 2015, 11:09
3
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 2 < 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.


a= 5^15 - 625^3 => 5^15 - (5^4)^3 => 5^15 - 5^12 = 5^12(5^3 - 1) = 5^12*124

124 = 31*4

a/x is integer

For condition of 2<p<x only 5 and 31 satisfies this


Hence answer is C

Originally posted by King407 on 15 May 2015, 10:31.
Last edited by King407 on 15 May 2015, 11:09, edited 1 time in total.
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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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New post 15 May 2015, 10:49
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a=5^15-625^3=5^15-5^12=5^12*124

Now a/x=integer and 124=4x31
Only number satisfying the given conditions are 5 and 31
C
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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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New post 15 May 2015, 11:04
1
Resolve a= (625^3)*(625^4)-625^3 = 625^3(625^4-1); so it can have two values one is 625^3 & (625^4-1) => multiple of 5 and 4

Given that a=+ int * X, therefore X can definitely have two values.

But it may have more than two as well, as it can be 5 * 25*625…*4. So here I’m confused. But still I will go with answer two, option C

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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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New post 15 May 2015, 12:52
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This is a tricky worded question and I think the answer is should be D not C...

Here is my reason :

The stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . The stem wants to say that X is a PRIME NUMBER . because any prime

Number has no factor grater than 1 and Itself .

On the other hand the stem says that X COULD get how many different number NOT MUST get different number ( this is very important issue )

AS our friends say, if we simplify Numerator more we can obtain : 5^12 ( 5^3-1) = 5^12 (124) = 5^12 (31*2*2) divided by x and we are told that this fraction is

An integer . so, X COULD Be ( not Must be) 5 , 31 ,or 2 !!! so , X could get 3 different values and answer is D.... :P :P :P

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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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New post 15 May 2015, 13:00
mehrdadtaheri92 wrote:
This is a tricky worded question and I think the answer is should be D not C...

Here is my reason :

The stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . The stem wants to say that X is a PRIME NUMBER . because any prime

Number has no factor grater than 1 and Itself .

On the other hand the stem says that X COULD get how many different number NOT MUST get different number ( this is very important issue )

AS our friends say, if we simplify Numerator more we can obtain : 5^12 ( 5^3-1) = 5^12 (124) = 5^12 (31*2*2) divided by x and we are told that this fraction is

An integer . so, X COULD Be ( not Must be) 5 , 31 ,or 2 !!! so , X could get 3 different values and answer is D.... :P :P :P

Best Regards,

Here we have the inequality 2<p<x and I assumed that x >2 is a given hence ignored 2. Otherwise we need not consider x >2 and can include 1 to the list and that would make 4 roots.
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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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New post 15 May 2015, 13:07
masoomdon wrote:
mehrdadtaheri92 wrote:
This is a tricky worded question and I think the answer is should be D not C...

Here is my reason :

The stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . The stem wants to say that X is a PRIME NUMBER . because any prime

Number has no factor grater than 1 and Itself .

On the other hand the stem says that X COULD get how many different number NOT MUST get different number ( this is very important issue )

AS our friends say, if we simplify Numerator more we can obtain : 5^12 ( 5^3-1) = 5^12 (124) = 5^12 (31*2*2) divided by x and we are told that this fraction is

An integer . so, X COULD Be ( not Must be) 5 , 31 ,or 2 !!! so , X could get 3 different values and answer is D.... :P :P :P

Best Regards,

Here we have the inequality 2<p<x and I assumed that x >2 is a given hence ignored 2. Otherwise we need not consider x >2 and can include 1 to the list and that would make 4 roots.[/quote

X can not be 1 since from the stem we should understand that X is a prime number and 1 is NOT a prime number . so 1 CAN NOT include in the possible roots,...
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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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New post 15 May 2015, 13:13
Its actually not mentioned that x is a prime number . from the inequality 2<p<x we assume that x has to be greater than 2 for the equality to make sense but if we forgo that then both 2 and 1 will be on the roots.
I still feel only numbers will be 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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New post 15 May 2015, 13:22
masoomdon wrote:
Its actually not mentioned that x is a prime number . from the inequality 2<p<x we assume that x has to be greater than 2 for the equality to make sense but if we forgo that then both 2 and 1 will be on the roots.
I still feel only numbers will be 5 and 31.




OK . May be you right. let wait till the OE and OA. But if you read the question stem more carefully, you will see that it says X has NO factor GRATER than 2 and less THAN x itself.

I think it means X is a prime number. Because any prime number has ONLY 1 and itself in its factor list...
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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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New post 15 May 2015, 13:29
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. For a prime number the inequality would be 2<=p<=x whereas here we have 2<p<x.anyways let's wait for the OA and OE
I remembered it just now so edited the post instead of adding one more post and hijacking the thread
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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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New post 15 May 2015, 16:28
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the answer is E
the question say that x does not has factore less than x itself and more than 2 which mean that x does not have factors in this range
a =(5^15-5^12)
=5^12(5^3-1)=5^12*124
so x could be 1, 5,2,13 all these numbers do not have factores less than the numbers itself and in the same time these factors are more than 2

In other word, the two condations sould not be find to say that the number is x

first condition is that the number has factors less than the number itself secondly the factors should be more than 2.

for example 2 have the factore 1 which is less than 2 itself this is the first does not match with x condition were x does not havs this point but 1 is not more than 2 so we can say that 2 could be x were 2 DOES NOT has a factor in this range.

finally we can not say that x should be prime number because if x is prime number then x will be DO NOT HAS FACTOR 1<P<X.
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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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New post 15 May 2015, 21:16
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 2 < p < x, then how many different values for x are possible?

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


Ans: D

Solution: a=5^12*31*4 (after generalizing it)
now a/x is an integer x must be a factor. so the question can have three different factors, which satisfy the condition 2< p <x
are 5, 31, 4
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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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New post 16 May 2015, 01:22
e is the answer x could be 2,4,31 & 5
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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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New post 18 May 2015, 07:11
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.


Sorry guys, it's 1 < p < x not 2 < p < x. Edited.
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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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New post 19 May 2015, 08:29
I just wonder why x can't be 1.
"1" does not violate the statement that "x is a positive integer such that it does NOT have a factor p such that 1 < p < x".
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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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Re: a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer  [#permalink]

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New post 19 May 2015, 08:45
Bunuel wrote:
VladimirKarpov wrote:
I just wonder why x can't be 1.
"1" does not violate the statement that "x is a positive integer such that it does NOT have a factor p such that 1 < p < x".



1 < p < x means that 1 < x.



But the statement gives us information that scheme "1<p<x" does not work here. I don't think we can interpret it as 1 < x.
I will make substitution in the statement: "1 is a positive integer such that it does NOT have a factor p such that 1<p<x". That's right - it does not have.
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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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New post 19 May 2015, 08:49
VladimirKarpov wrote:
Bunuel wrote:
VladimirKarpov wrote:
I just wonder why x can't be 1.
"1" does not violate the statement that "x is a positive integer such that it does NOT have a factor p such that 1 < p < x".



1 < p < x means that 1 < x.



But the statement gives us information that scheme "1<p<x" does not work here. I don't think we can interpret it as 1 < x.
I will make substitution in the statement: "1 is a positive integer such that it does NOT have a factor p such that 1<p<x". That's right - it does not have.


Edited the question to avoid ambiguity.
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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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New post 03 Dec 2016, 08:43
a = 5^15 - 5^12 (As 625 = 5^4 and 5^4^3 = 5^12)
a= 5^12(5^3 - 1)
a = 5^12(124)...124 = 2*2*31

So, 2, 5, 31 are prime numbers (Not having divisors other than 1 and self)

So, D is the correct answer
Re: a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer &nbs [#permalink] 03 Dec 2016, 08:43

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