Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43892

a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 02:50
Question Stats:
48% (01:23) correct 52% (01:34) wrong based on 334 sessions
HideShow timer Statistics



Manager
Joined: 03 Sep 2014
Posts: 75
Concentration: Marketing, Healthcare
Schools: Kellogg 1YR '17, Booth '16, McCombs '18, Tepper '18, INSEAD Jan '17, ISB '17, NUS '18, IIMA , IIMB, IIMC , IIML '15

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 09:31
2
This post received KUDOS
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
Last edited by King407 on 15 May 2015, 10:09, edited 1 time in total.



Manager
Joined: 07 Feb 2015
Posts: 50

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 09:49
1
This post received KUDOS
a=5^15625^3=5^155^12=5^12*124 Now a/x=integer and 124=4x31 Only number satisfying the given conditions are 5 and 31 C
_________________
Kudos if it helped you in any way



Manager
Joined: 26 Dec 2012
Posts: 148
Location: United States
Concentration: Technology, Social Entrepreneurship
WE: Information Technology (Computer Software)

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 10:04
1
This post received KUDOS
Resolve a= (625^3)*(625^4)625^3 = 625^3(625^41); so it can have two values one is 625^3 & (625^41) => 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
Thanks,



Manager
Joined: 13 Dec 2013
Posts: 59
Location: Iran (Islamic Republic of)

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 11:52
2
This post received KUDOS
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 ( 5^31) = 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.... Best Regards,



Manager
Joined: 07 Feb 2015
Posts: 50

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 12:00
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 ( 5^31) = 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.... 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.
_________________
Kudos if it helped you in any way



Manager
Joined: 13 Dec 2013
Posts: 59
Location: Iran (Islamic Republic of)

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 12:07
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 ( 5^31) = 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.... 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,...



Manager
Joined: 07 Feb 2015
Posts: 50

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 12:13
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.
_________________
Kudos if it helped you in any way



Manager
Joined: 13 Dec 2013
Posts: 59
Location: Iran (Islamic Republic of)

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 12:22
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...



Manager
Joined: 07 Feb 2015
Posts: 50

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 12:29
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
_________________
Kudos if it helped you in any way



Manager
Status: Kitchener
Joined: 03 Oct 2013
Posts: 95
Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 15:28
1
This post received KUDOS
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 =(5^155^12) =5^12(5^31)=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 2In other word, the two condations sould not be find to say that the number is xfirst 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.
_________________
Click +1 Kudos if my post helped



Manager
Joined: 21 Jan 2015
Posts: 149
Location: India
Concentration: Strategy, Marketing
WE: Marketing (Consumer Products)

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
15 May 2015, 20:16
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
_________________
 The Mind is Everything, What we Think we Become. Kudos will encourage many others, like me. Please Give Kudos !! Thanks



Manager
Joined: 05 Mar 2013
Posts: 75
Location: United States
Schools: Wharton '17, Kellogg '17, Booth '17, Ross '17, Haas '17, Tuck '17, Yale '17, Duke '17, Anderson '17, Johnson '17, Kelley '17, McDonough '17, Goizueta '17
GPA: 3.56
WE: Marketing (Telecommunications)

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
16 May 2015, 00:22
e is the answer x could be 2,4,31 & 5



Math Expert
Joined: 02 Sep 2009
Posts: 43892

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
18 May 2015, 06:10
1
This post received KUDOS
Expert's post
6
This post was BOOKMARKED
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, \(a = 5^{15}  625^3=5^{15}  5^{12}=5^{12}(5^31)=5^{12}*124=2^2*5^{12}*31\). Finally, for a/x to be an integer where x is a prime, x can take 3 values: 2, 5, or 31. Answer: D.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 43892

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
18 May 2015, 06:11



Intern
Joined: 24 Apr 2013
Posts: 16
Location: Russian Federation

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
19 May 2015, 07:29
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".



Math Expert
Joined: 02 Sep 2009
Posts: 43892

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
19 May 2015, 07:33



Intern
Joined: 24 Apr 2013
Posts: 16
Location: Russian Federation

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
19 May 2015, 07:45
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.



Math Expert
Joined: 02 Sep 2009
Posts: 43892

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
19 May 2015, 07:49



NonHuman User
Joined: 09 Sep 2013
Posts: 13786

Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer [#permalink]
Show Tags
01 Dec 2016, 13:59
Hello from the GMAT Club BumpBot! 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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: a = 5^15  625^3 and a/x is an integer, where x is a positive integer
[#permalink]
01 Dec 2016, 13:59



Go to page
1 2
Next
[ 23 posts ]



