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

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 \(n^{st}\)

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

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.

chetan2u
what is the rule that allows this?: s^st is a factor, so s^t will also be a factor..

I have added few more details for better understanding, so please check it again.
Should clear your query.

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

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!

(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?

This is the algebraic approach I figured out to solve this question.

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.

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