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Posts: 63
Updated on: 13 Jun 2021, 02:00
Originally posted by [email protected] on 08 May 2016, 20:14.
Last edited by Bunuel on 13 Jun 2021, 02:00, edited 2 times in total.
Last edited by Bunuel on 13 Jun 2021, 02:00, edited 2 times in total.
Edited the question.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
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Question Stats:
68% (01:53) correct
32%
(01:52)
wrong
based on 3820
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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 \(n^{st}\)?
I. \(s^t\)
II. \((st)^2\)
III. \(s + t\)
(A) None
(B) I only
(C) II only
(D) III only
(E) I and II only
I. \(s^t\)
II. \((st)^2\)
III. \(s + t\)
(A) None
(B) I only
(C) II only
(D) III only
(E) I and II only
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Abhishek009
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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.
_________________
If s=ax and b=bx, where a, b and x are integers, and as s and t are factors of n, we can write n=LCM(st)*y=abxy
\(n^{st}\) will have factors: \(x^{st}, \ \ s^{st}, \ \ and \ \ t^{st}\)
lets see the choices..
\(1) s^t\)
\(s^{st}\) is a factor, so \(s^t\) and \(s^s\) will also be a factor.. YES
\(2) (st)^2\)
\((s)^{st}\) and \((t)^{st}\) are factors and we know that both s and t are >1. Thus least value of both s and t is 2.
So, the min value of \(n^{st}\) will be \(n^4=(abxy)^4=a^4*b^4*x^4*y^4=a^2*(ax)^2*b^2*(bx)^2*b^2*y^4=a^2*s^2*t^2*y^4=a^2*(st)^2*y^4\).
Thus, \((st)^2\) will always be a factor of \((n)^4\) or \(n^{st}\).. YES
\(3) s + t\)
we do not know what are the factors of s and t...
Say y=5, x=1, a=2, and b=3, that is s=2 and t=3, then n=5*2*3=30. 2+3 or 5 is a factor of \(30^{2*3}\)
But say y=5, x=1, a=2, and b=3, that is s=2 and t=3, then n=6*2*3=36. 2+3 or 5 is not a factor of \(36^{2*3}\)
NOT necessary
Only (1) and (2) must be true.
E
Responding to a PM
Adding my take on substituting some values for s and t.
Always be careful on taking values in MUST be true questions.
You may have just taken some value that fits in for the values taken but may not fit in for other values.
For example: if you take s=2 and t=6, then n=LCM (2,6)*y=6y
Option III. s+t=2+6=8, which will be a factor of (6y)^12=2^12*3^12*y^12.
Or
If you take s=t, option III will be true irrespective of values of s and t.
_________________
General Discussion
chetan2u
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!!
badboson
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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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^(st-4))*(ab^2)
This is always an integer as st>=4
3) is clearly not a factor
Thus option E
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^(st-4))*(ab^2)
This is always an integer as st>=4
3) is clearly not a factor
Thus option E
madgabriel
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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? \(\rightarrow\) \(s^{pst-t}t^{qst}=s^{t(ps-1)}t^{qst}\). 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 \(t(ps-1)\) will always be positive and therefore \(s^{t(ps-1)}t^{qst}\) will always be an integer. This statement must be true.
II. Is \(\frac{s^{pst}t^{qst}}{(st)^{2}}\) an integer? \(\rightarrow\) \(s^{pst-2}t^{qst-2}\). 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 \(0\), \(pst-2\) and \(qst-2\) will always be positive and therefore \(s^{pst-2}t^{qst-2}\) 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.
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? \(\rightarrow\) \(s^{pst-t}t^{qst}=s^{t(ps-1)}t^{qst}\). 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 \(t(ps-1)\) will always be positive and therefore \(s^{t(ps-1)}t^{qst}\) will always be an integer. This statement must be true.
II. Is \(\frac{s^{pst}t^{qst}}{(st)^{2}}\) an integer? \(\rightarrow\) \(s^{pst-2}t^{qst-2}\). 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 \(0\), \(pst-2\) and \(qst-2\) will always be positive and therefore \(s^{pst-2}t^{qst-2}\) 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.
chetan2u
what is the rule that allows this?: s^st is a factor, so s^t will also be a factor..
what is the rule that allows this?: s^st is a factor, so s^t will also be a factor..
chetan2u
dc2880
I have added few more details for better understanding, so please check it again.
Should clear your query.
_________________
chetan2u
thank you for your reply. for "\(s^{st}\) is a factor, so \(s^t\) and \(s^s\) will also be a factor.. YES"
wouldn't \(s^{st}\) factor out to \((s^{t})^s\), not \(s^t\) * \(s^s\)?
dc2880
Yes, \(s^{st}=(s^s)^t\), so \(s^s \) is a factor.
Similarly, \(s^{st}=(s^t)^s\), so \(s^t \) is a factor.
Thus both s^t and s^s are factors.
_________________
VeritasKarishma Bunuel BrentGMATPrepNow egmat
I tested some values, and got the right answer. But I wasn't totally sure if these 2 options are going to be valid for ANY set of numbers.
Could you pls explain it in an algebraic way, or maybe in some way using which I could be totally sure that the numbers will work everytime?
Thankyou in Advance!
Posts: 7
(I) If s^t is a factor of n^(st), then n^(st) / s^t is an integer. From the q we know n/s and n/t are integers (as s and t are factors of n).
Then n^(st) / s^t = n^t * n^(-t) * n^(st) / s^t = (n/s)^t * n^(st)*n^(-t) = integer * n^(t(s-1)) = integer, since t and s are > 1 (so s-1 is at least 1)
(II) Is n^(st) / (st)^2 an integer? We know n/s * n/t = n^2/(st) is an integer. So is n^4 / (st)^2
Then n^(st) / (st)^2 = n^4 / (st)^2 * n^(st) * n^-4 = integer * n^(st-4) = integer, since s and t are > 1, so st>=4, so n^(st-4) is at least 1
At this point the correct answer must be E (I&II are true). There is no option for I&II&III true. Not sure how this could be done in 2 mins though! Anyone see errors with my solution?
Then n^(st) / s^t = n^t * n^(-t) * n^(st) / s^t = (n/s)^t * n^(st)*n^(-t) = integer * n^(t(s-1)) = integer, since t and s are > 1 (so s-1 is at least 1)
(II) Is n^(st) / (st)^2 an integer? We know n/s * n/t = n^2/(st) is an integer. So is n^4 / (st)^2
Then n^(st) / (st)^2 = n^4 / (st)^2 * n^(st) * n^-4 = integer * n^(st-4) = integer, since s and t are > 1, so st>=4, so n^(st-4) is at least 1
At this point the correct answer must be E (I&II are true). There is no option for I&II&III true. Not sure how this could be done in 2 mins though! Anyone see errors with my solution?
This is the algebraic approach I figured out to solve this question.
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Abhishek009
Abhishek009 KarishmaB
Is it recommended to use numbers in these kind of problems with factors?
Although factors, divisibility etc. are typically fun and logical, I get bogged down given the time restriction. I got it wrong in the GMAT prep test, but when I solved it later with numbers, it was easy. And I found this solution quite solvable within 2 mins as well. Thank you for any words of wisdom.
Engineer1
Yes & No.
When I read it, I used numbers to make sense of what it is saying. But after I realised what it is saying, I didn't want to continue with using numbers for variables because there were 3 of them. That just means a lot of confusion.
So I went back to the question and once again evaluated what it was saying.
s and t are factors of n so I may have something similar to
\((st...)^{st}\) (Of course it is possible that s = 4, t = 6 and n = 12 and hence this format may not hold. )
\(s^t\)
will be a factor
\((st)^2\)
will be a factor. Both s and t are integers greater than 1 so st will be greater than 2. Hence (st)^2 will be a factor of (st)^st
s + t
Not necessary. st may not have s+t as a factor. e.g. s=2 and t = 3, then 6^6 has no 5.
Answer (E)
Check out this video on exponents: https://youtu.be/ibDqnatAMG8
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Funny enough, I picked the exact set of numbers for my example.
Abhishek009
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