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If s and t are integers greater than 1 and each is a factor of the int Updated on: 13 Jun 2021, 03:00
Originally posted by [email protected] on 08 May 2016, 21:14.
Last edited by Bunuel on 13 Jun 2021, 03:00, edited 2 times in total.
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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
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E
Solving this question helps. Taking a timed set of similar questions in GMAT Club Forum Quiz → is even better.
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If s and t are integers greater than 1 and each is a factor of the int 15 May 2016, 10:17
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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
Show more

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. :P
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If s and t are integers greater than 1 and each is a factor of the int Updated on: 12 Jun 2021, 20:56
Originally posted by chetan2u on 08 May 2016, 21:24.
Last edited by chetan2u on 12 Jun 2021, 20:56, edited 1 time in total.
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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
Show more

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.
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If s and t are integers greater than 1 and each is a factor of the int 15 May 2016, 10:37
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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

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
Show more

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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If s and t are integers greater than 1 and each is a factor of the int 15 May 2016, 10:52
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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

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!!
Show more

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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If s and t are integers greater than 1 and each is a factor of the int 14 Feb 2017, 03:50
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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
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If s and t are integers greater than 1 and each is a factor of the int 27 May 2017, 20:52
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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.
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If s and t are integers greater than 1 and each is a factor of the int 12 Jun 2021, 14:17
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chetan2u
what is the rule that allows this?: s^st is a factor, so s^t will also be a factor..

chetan2u
[email protected]
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
Show more
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If s and t are integers greater than 1 and each is a factor of the int 12 Jun 2021, 20:44
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dc2880
chetan2u
what is the rule that allows this?: s^st is a factor, so s^t will also be a factor..

chetan2u
[email protected]
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
Show more

I have added few more details for better understanding, so please check it again.
Should clear your query.
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If s and t are integers greater than 1 and each is a factor of the int 12 Jun 2021, 23:50
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chetan2u
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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

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.
Show more

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 \(s^t\) * \(s^s\)?
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If s and t are integers greater than 1 and each is a factor of the int 13 Jun 2021, 01:57
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chetan2u
[email protected]
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

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.

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 \(s^t\) * \(s^s\)?
Show more

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.
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If s and t are integers greater than 1 and each is a factor of the int 06 Nov 2021, 11:18
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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
Show more

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!
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If s and t are integers greater than 1 and each is a factor of the int 22 Jan 2022, 08:41
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(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?
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If s and t are integers greater than 1 and each is a factor of the int 08 Feb 2023, 11:46
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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
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This is the algebraic approach I figured out to solve this question.
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Engineer1
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If s and t are integers greater than 1 and each is a factor of the int 13 Jan 2024, 21:52
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Abhishek009
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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

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. :P
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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.
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If s and t are integers greater than 1 and each is a factor of the int 14 Jan 2024, 02:51
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Engineer1
Abhishek009
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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

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. :P

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.
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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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If s and t are integers greater than 1 and each is a factor of the int 22 Sep 2024, 17:19
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Funny enough, I picked the exact set of numbers for my example.
Abhishek009
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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

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. :P
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If s and t are integers greater than 1 and each is a factor of the int 28 Sep 2025, 23:57
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