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If s and t are integers greater than 1 and each is a factor of the int
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Updated on: 22 Oct 2016, 03:32
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63% (00:54) correct 37% (00:54) wrong based on 763 sessions
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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
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Re: If s and t are integers greater than 1 and each is a factor of the int
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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
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Re: If s and t are integers greater than 1 and each is a factor of the int
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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.
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Re: If s and t are integers greater than 1 and each is a factor of the int
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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!!



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Re: If s and t are integers greater than 1 and each is a factor of the int
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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 \(n^{st}\)
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Re: If s and t are integers greater than 1 and each is a factor of the int
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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



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Re: If s and t are integers greater than 1 and each is a factor of the int
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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.



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Re: If s and t are integers greater than 1 and each is a factor of the int
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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?



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Re: If s and t are integers greater than 1 and each is a factor of the int
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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?



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Re: If s and t are integers greater than 1 and each is a factor of the int
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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



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Re: If s and t are integers greater than 1 and each is a factor of the int
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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
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If s and t are integers greater than 1 and each is a factor of the int
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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



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If s and t are integers greater than 1 and each is a factor of the int
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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.



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Re: If s and t are integers greater than 1 and each is a factor of the int
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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.




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22 Nov 2017, 08:20






