Definitely prime factorization! First, it's easiest to see how you can factor if you write both exponential expressions with the same base:

4^17 - 2^28 = (2^2)^17 - 2^28 = 2^34 - 2^28

Both 2^34 and 2^28 are divisible by 2^28, so we can factor that out:

2^34 - 2^28 = 2^28 * ( 2^6 - 1)

Now, the expression in brackets is small enough to calculate:

2^28 * ( 2^6 - 1) = 2^28 * (64 - 1) = 2^28 * 63 = 2^28 * 3^2 * 7

Once you get this far, any question about divisibility is easy to answer. For example, what's the largest prime divisor? It's 7.

vinviper1 wrote:I need to work on my prime factorization skills. Is that simply breaking down the number until you get some primes?

Yes, that's essentially the idea- really, it's dividing up the number until you get *only* primes. If you need to prime factorize, just start dividing any way you like, and keep going until all you have are primes. For example, start with 90:

90 = 9*10 = 3*3*2*5

You don't need to start with 9*10; you'll get the same answer no matter how you begin. So you could have instead done;

90 = 2*45 = 2*9*5 = 2*3*3*5

which is the same as the above, just in a different order.

Number theorists normally write prime factorizations by using exponents, and normally write the primes in increasing order, so the above prime factorization would normally be written 2*3^2*5.

One of the most important facts in all of mathematics is that prime factorizations are 'unique'. If you want to multiply primes together to get 90, for example, you must multiply one 2, one 5, and two 3s. There's no other way to do it. It is because of this that prime factorizations are so useful; if you know 90 = 2*3^2*5, you can see all of the divisors of 90. As a simple example, you can see immediately that the only primes which are divisors of 90 are 2, 3, and 5.

Prime factorization is an amazingly useful technique, especially on harder GMAT questions that mention 'divisibility' or 'divisors'. Experience helps a lot here, though. Not only do you need to know how to find a prime factorization, but often you need to know what to do after you've found it. That part is often different from question to question, and it's very useful to see many different examples. The question you've posted is a great start, and a good example of a fairly difficult GMAT question on prime factorization.

nifoui7 wrote: Hi Ian,

I know this post is already quite old, but I've been trying to understand prime factorization and I'm stuck here with your explanation.

I'm lost when you go from this 2^34 - 2^28 to this 2^28 * ( 2^6 - 1)

Can you explain plz?

Thx a lot!

Just to add to the explanation above: the factorization we're doing here is one you probably will have done many times in algebra questions. When you see something like x^5 - x^3, we'd almost always want to factor out the term with the smaller power: x^5 - x^3 = x^3(x^2 - 1). We're using the same type of factorization in the question above - it's just that we have a base of 2 instead of x.