Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 17 Sep 2015
Posts: 83

If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
Updated on: 22 Oct 2016, 03:32
Question Stats:
62% (01:49) correct 38% (01:45) wrong based on 1358 sessions
HideShow timer Statistics
If \(s\) and \(t\) are integers greater than \(1\) and each is a factor of the integer \(n\), which of the following must be a factor of \(n^{st}\)? 1) \(s^t\) 2) \((st)^2\) 3) \(s + t\) A) None B) 1 only C) 2 only D) 3 only E) 1 and 2
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
You have to dig deep and find out what it takes to reshuffle the cards life dealt you




Math Expert
Joined: 02 Aug 2009
Posts: 8327

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
08 May 2016, 21:24
aniketm.87@gmail.com wrote: if s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n^st?
1) s^t
2) (st)^2
3) s + t
A) None
B) 1 only
C) 2 only
D) 3 only
E) 1 and 2 lets see the choices.. \(1) s^t\) s^st is a factor, so s^t will also be a factor.. YES \(2) (st)^2\) as s and t are >1, in n^st, st will be atleast 4, as minimum value of s and t is 2.. so n^st or min value n^4 = st^4, thus st^2 will always be a factor of n^st.. YES \(3) s + t\) we do not know what are the factors of s and t... NOT necessary E
_________________




Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4868
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
15 May 2016, 10:17
aniketm.87@gmail.com wrote: If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n^st?
1) s^t
2) (st)^2
3) s + t
A) None
B) 1 only
C) 2 only
D) 3 only
E) 1 and 2 Let s =2 and t = 3 and n =6 { Where s and t are integers greater than 1 and each is a factor of the integer 6 }\(n^{st}\) = \(6^6\) =>\({2^6}{3^6}\) Now check for the options  1) \(s^t\)= \(2^3\) ( Can be a factor of \({2^6}{3^6}\) )2) \((st)^2\) = \({s^2}{t^2}\) => \({2^2}{3^2}\) ( Can be a factor of \({2^6}{3^6}\) )\(3)\) \(s\) \(+\) \(t\) \(=\) \(2\) \(+\) \(3\) = \(5\) ( Can not be a factor of \({2^6}{3^6}\) )Hence only option E) 1 and 2 follows.
_________________




Intern
Joined: 15 Jan 2016
Posts: 17

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
15 May 2016, 10:37
chetan2u wrote: aniketm.87@gmail.com wrote: if s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n^st?
1) s^t
2) (st)^2
3) s + t
A) None
B) 1 only
C) 2 only
D) 3 only
E) 1 and 2 lets see the choices.. \(1) s^t\) s^st is a factor, so s^t will also be a factor.. YES \(2) (st)^2\) as s and t are >1, in n^st, st will be atleast 4, as minimum value of s and t is 2.. so n^st or min value n^4 = st^4, thus st^2 will always be a factor of n^st.. YES \(3) s + t\) we do not know what are the factors of s and t... NOT necessary E Hi Chetan2U, From the info given in the problem, I do not think we can write n = x * s * t, where x is integer. For example, if n = 18, then possible value for s & t can be 9 & 6 respectively. Then \(n^s^t\) cannot be written as \((xst)^s^t\). So i am still not clear how we can conclude '2' is an answer without putting individual values. Please let me know if i am missing something. Thanks in advance!!



Math Expert
Joined: 02 Aug 2009
Posts: 8327

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
15 May 2016, 10:52
badboson wrote: aniketm.87@gmail.com wrote: if s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n^st?
1) s^t
2) (st)^2
3) s + t
A) None
B) 1 only
C) 2 only
D) 3 only
E) 1 and 2 Hi Chetan2U, From the info given in the problem, I do not think we can write n = x * s * t, where x is integer.For example, if n = 18, then possible value for s & t can be 9 & 6 respectively. Then \(n^s^t\) cannot be written as \((xst)^s^t\). So i am still not clear how we can conclude '2' is an answer without putting individual values. Please let me know if i am missing something. Thanks in advance!! Hi, yes you are correct on the highlighted portion.. we are looking for factor of \(n^{st}\), where s and t are factor of n.. 2. \((st)^2\)... we know s and t are factors of n, and they are >1.. lets assume the max value of s and t as n.. so the term = \(n^{st} = n^{n^2}\)..(i) and \((st)^2 = (n*n)^2 = (n^2)^2 = n^4\)..(ii) Now\(n^{n^2} >= n^{2^2} or n^4\).. because if s and t are integers >1 and are factor of n, then n is atleast 2.. so clearly \((st)^2\) is a factor of \(n^{st}\)
_________________



Manager
Status: 2 months to go
Joined: 11 Oct 2015
Posts: 100
GPA: 3.8

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
31 Jul 2016, 12:10
Abhishek009 wrote: aniketm.87@gmail.com wrote: If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n^st?
1) s^t
2) (st)^2
3) s + t
A) None
B) 1 only
C) 2 only
D) 3 only
E) 1 and 2 Let s =2 and t = 3 and n =6 { Where s and t are integers greater than 1 and each is a factor of the integer 6 }\(n^{st}\) = \(6^6\) =>\({2^6}{3^6}\) Now check for the options  1) \(s^t\)= \(2^3\) ( Can be a factor of \({2^6}{3^6}\) )2) \((st)^2\) = \({s^2}{t^2}\) => \({2^2}{3^2}\) ( Can be a factor of \({2^6}{3^6}\) )\(3)\) \(s\) \(+\) \(t\) \(=\) \(2\) \(+\) \(3\) = \(5\) ( Can not be a factor of \({2^6}{3^6}\) )Hence only option E) 1 and 2 follows. Hey Abhishek, a quick question. Given that I solved the problem the same way (tried 2 & 2 and 3 & 5), on test day would you have tried other cases or would you have trusted your first result? In the second case, why would you have trusted it? Abhishek009Den ps. Chetan I clearly value your answer as much. chetan2u



Current Student
Joined: 28 Nov 2014
Posts: 812
Concentration: Strategy
GPA: 3.71

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
26 Aug 2016, 14:09
Abhishek009 chetan2u Even I want to know as to how can we believe that the example we have chosen will give us an exhaustive list and we will not have to check any further cases.



Intern
Joined: 27 Mar 2016
Posts: 23
Location: India
Concentration: Entrepreneurship, Social Entrepreneurship
WE: General Management (Consumer Products)

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
06 Oct 2016, 07:47
aniketm.87@gmail.com wrote: If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n^st?
1) s^t
2) (st)^2
3) s + t
A) None
B) 1 only
C) 2 only
D) 3 only
E) 1 and 2 What should be the Best Method to solve this kind of question "WHICH OF THE FOLLOWING MUST BE/ATLEAST"? I. II. III None Will test cases be the best method when the clock is running out? And will our intention be negating the Question stem?



Current Student
Joined: 28 Nov 2014
Posts: 812
Concentration: Strategy
GPA: 3.71

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
07 Oct 2016, 02:13
Can anyone confirm whether taking cases will yield a definite answer? How do we know if we have checked all the cases?



Retired Moderator
Joined: 05 Jul 2006
Posts: 1373

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
10 Jan 2017, 03:38
Keats wrote: Can anyone confirm whether taking cases will yield a definite answer? How do we know if we have checked all the cases? N^st = k^st * (st)^st = k^st * s^st * t^st = k^st * (s^t)^s * t^st key is knowing each of s,t >=2 ( nothing in the problem mentioned they are different integers) now divide by each given answer option 1 k^st * (s^t)^s * t^st / s^t .... must yield an integer 2 k^st * (st)^st / (st)^2 = k^st * (st)^st2 ... thus question becomes is it a must that st>=2 the answer is yes since minimum value of ST ( if we assumed they are equal to their possible minimum = 2*2 = 4 ) ... hope this helps



Manager
Joined: 25 Mar 2013
Posts: 221
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
10 Jan 2017, 09:14
s > 1 , t > 1 s,t factors of n sxt also will be a factor of n Take s = 2, t =3 , n = 6 1, 2^3 True 2, (2x3)^2 True 3, 3+2 = 5 False
E



Intern
Joined: 08 Dec 2016
Posts: 6
Location: India
GPA: 3.6

If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
14 Feb 2017, 03:50
We can check this by
n = a*s where a>=1 (s is a factor of n) n = b*t where b>=1 (t is a factor of n)
1) n^st = (a*s)^st = a^st * s^st = a^st*s^s*s^t since a^st*s^s is a positive integer thus s^t is a factor
2) for st^2 to be a factor n^st/st^2 should be an integer always n^st/st^2 = n^st / ((n/a)*(n/b))^2 = n^st/(n^2/ab)^2 = (n^(st4))*(ab^2) This is always an integer as st>=4
3) is clearly not a factor
Thus option E



Current Student
Joined: 27 May 2015
Posts: 12
Location: Venezuela
GPA: 3.76

If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
27 May 2017, 20:52
Generally, this kind of "must be true" questions are ideal for number picking. Since it "must be true", any numbers you pick (provided you comply with the conditions mentioned in the question stem) should be valid.
However, we can try an algebraic approach as well (for those, like me, who don't like the number picking approach).
We are told that \(s\) and \(t\) are integers greater than \(1\) and that each one of them is a factor of the integer \(n\). We can take this information as \(n=s^{p}t^{q}\) (regardless of other factors \(n\) could have), where \(p\) and \(q\) are integers greater than \(0\) (since we need \(s\) and \(t\) to be factors of \(n\)). Therefore, \(n^{st}\) is the same as \(s^{pst}t^{qst}\)
Now we can go statement by statement:
I. Is \(\frac{s^{pst}t^{qst}}{s^{t}}\) an integer? \(\rightarrow\) \(s^{pstt}t^{qst}=s^{t(ps1)}t^{qst}\). It's important to notice that since \(t\) and \(s\) are greater than \(1\) (according to the question stem) and since \(p\) must be greater than \(0\), then \(t(ps1)\) will always be positive and therefore \(s^{t(ps1)}t^{qst}\) will always be an integer. This statement must be true.
II. Is \(\frac{s^{pst}t^{qst}}{(st)^{2}}\) an integer? \(\rightarrow\) \(s^{pst2}t^{qst2}\). Again, it's important to notice that since \(t\) and \(s\) are greater than \(1\) (according to the question stem) and since \(p\) and \(q\) must be greater than \(0\), \(pst2\) and \(qst2\) will always be positive and therefore \(s^{pst2}t^{qst2}\) will always be an integer. This statement must be true.
III. Is \(\frac{s^{pst}t^{qst}}{s+t}\) an integer? \(\rightarrow\) Well, in this case we know that a sum of positive factors is never a factor of the multiplication of such factors. You can try this: \(\frac{(6)(5)}{6+5}=\frac{30}{11}\), which is clearly not an integer. This statement is not true.
Correct answer is option E.
Hope this helps.



Intern
Joined: 27 Oct 2017
Posts: 8
Location: India
GPA: 4

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
22 Nov 2017, 08:20
As far as the confusion with (st)^2. It's quite simple to understand that it must be a factor. Consider the following: s & t are greater than 1. Thus s and t have to be at least 2 each. Say s=2+u and t=2+v. (n)^st = n^((2+u)(2+v)) = (n^4)*n^p for some p. Hence, (n)^st = (n^4)*q (for q=n^p). Now s and t are factors of n. Hence, n=ys and n=zt for some integers y and z. Thus, (n)^st = (n^4)*q = (n^2)*(n^2)*q = ((ys)^2)*((zt)^2)*q = ((st)^2)*r for some r. Hence, (st)^2 must be a factor of n^st I'm sorry for the complicated notations.



Manager
Joined: 22 Jan 2018
Posts: 51

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
18 Dec 2018, 07:43
Can an expert explain this question. I have read the explanations, but am not very clear.



Intern
Joined: 29 Nov 2017
Posts: 15
GMAT 1: 700 Q49 V38 GMAT 2: 710 Q48 V39

Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
Show Tags
21 Jan 2019, 06:06
madgabriel wrote: Generally, this kind of "must be true" questions are ideal for number picking. Since it "must be true", any numbers you pick (provided you comply with the conditions mentioned in the question stem) should be valid.
That approach is optimal for "can be" questions, but not for "must be". You could have multiple right numbers picked, but if you missed a single one that doesn't work, then the "must" falls apart. madgabriel wrote: We are told that \(s\) and \(t\) are integers greater than \(1\) and that each one of them is a factor of the integer \(n\). We can take this information as \(n=s^{p}t^{q}\)
That seems wrong to me. It would work if \(s\) and \(t\) were prime factors, but the QS says they are just factors. The numbers could be 6 and 9, and n=18. Obviously \(18≠6^{p}9^{q}\) where p&q are integers greater than 0.




Re: If s and t are integers greater than 1 and each is a factor of the int
[#permalink]
21 Jan 2019, 06:06






