If n is an integer greater than 1 is 3^n-2^n divisible by 35

well I have been preparing for only a month now so am probably not the best person to answer this. Going by the problems I have found on this forum it was slightly on the tougher side.

But then again wont do anyone any harm to remember one simple rule

Very very good question
U should give more time for the others

+1 for both of you for 750+ level question and solution...

Thanks GMAT TIGER, but frankly speaking when I was composing this question I didn't think that it would be ranked as 750+.

Is it because the rule needed to deal with this problem is not widely known?

The question is perfect. It looks tough but solution is simple if you persue the right approach. Thats the characterstics of 750+ question, which are not essentially hard/difficult.....

Great question! I hope you can add more of these! I've seen a few of these questions but at lower level of difficulty. Usually the divisor is a factor of (a-b) or (a + b) where a, b are the base. So, it is easier to relate. Having 35 as divisor stumped me as I couldn't relate it to 3 and 2. I noticed that there are not a lot of discussion about this type of question either on the net or in the gmat book. So, I'm really glad to find it! Hope to see more!!!!Kudos for all your efforts. Cant tell you how much it helps..since I dont have access to any gmat books.

its a indeed a great quality question.

What is the reasoning behind these rules? (understanding makes it easier to remember)

I really doubt you need this.

RULE: for \(x^n-y^n\):
\(x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+x^{n-4}y^3+...+xy^{n-2}+y^{n-1})\)

So, \(x^n-y^n\) is always divisible by \(x-y\).

Now, when \(n=even\), then the second multiple will have n, so even terms and
\((x^{n-1}+x^{n-2}y+x^{n-3}y^2+x^{n-4}y^3+...+xy^{n-2}+y^{n-1})=\)
\(=(x^{n-2}(x+y)+x^{n-4}y^2(x+y)+...+y^{n-2}(x+y))=(x+y)(x^{n-2}+x^{n-4}y^2+...+y^{n-2})\)
So, \(x^n-y^n\) is also divisible by \(x+y\) when \(n\) is even.

Consider the following examples:
\(x^2-y^2=(x-y)(x+y)\): divisible by both \(x-y\) and \(x+y\);
\(x^3-y^3=(x-y)(x^2+xy+y^2)\): divisible by \(x-y\).

RULE: for \(x^n+y^n\):

When \(n=odd\) then:
\(x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^2-x^{n-4}y^3+...-xy^{n-2}+y^{n-1})\)
So, when \(n=odd\) then \(x^n+y^n\) is divisible by \(x+y\).

When \(n=even\) then \(x^n+y^n\) is NOT divisible by either \(x+y\) or \(x-y\).

Consider the following example:
\(x^3+y^3=(x+y)(x^2-xy+y^2)\): divisible by \(x+y\).

Thanks!
I hope this does not appear on GMAT too often.

Thanks. Corrected the typo.

hmmm. this is the kind of questions which make the prep interesting for anyone.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

I just wanted to show another method to prove the above. Perfect solution is already given by everyone.
+1 for the sum.
For any function f(x), if x=a makes f(a)=0, then (x-a) is a factor of the given function.

This is something we all know. It has just been put in words by me.To illustrate that, consider the given example :

f(x) =\(x^2-2x+1\) ; for x=1, f(1) = 0. Thus, (x-1) is a factor of f(x),and as everyone would have recognised,we all know f(x) = \((x-1)^2\).

I.Function of the form f(x) = \(x^n+y^n\) will have (x+y) as a factor, if and only if for x=-y, f(-y) = 0.
Thus, replacing the value of x in the above function, we get : \((-y)^n+y^n \to\) Notice that this will be zero , only if n is odd.
\(x^n+y^n\) will always have (x+y) as a factor, if n = odd.

II. Function of the form f(x) = \(x^n-y^n\). Just as above, for (x+y) to be a factor, for x = -y,f(-y) = 0

Again, by replacing f(-y) = \((-y)^n-y^n \to\) This will be zero only if n is even.

\(x^n-y^n\) will always have (x+y) as a factor, if n = even.

Again, for (x-y) to be a factor, for x = y, f(y) = 0.

f(y) = \(y^n-y^n\) = 0.

\(x^n-y^n\) will always have (x-y) as a factor, if n = odd/even.

very good question... thanks...

Good rules to remember but I think there is a typo above ,When you are explaining the logic for even case I think you meant to write
So, \(x^n-y^n\) is also divisible by \(x+y\) when \(n\) is even and not \(x-y\).

Good rules to keep in mind.

Thank you. Edited. +1.