Could someone show how to work the following?

What is the highest prime factor of

4^17 - 2^28

a. 2 b. 3 C. 5 D. 7 E. 11

Walter,

In many exponent problems involving different bases, you can convert one of the bases so that they end up the same. For example, in this problem, we can change 4^17 into (2^2)^17 = 2^34

So, we end up with:

2^34 - 2^28

Now, we can factor out a 2^28, so we have:

2^28(2^6 - 1)

We can simplify 2^6 - 1, because we can actually calculate 2^6 pretty easily.

2^6 - 1 = 64 -1 = 63

So, now we have:

2^28(63)

The greatest prime factor of 2^28 is 2. The greatest prime factor of 63 is 7. So, the answer to the problem is 7.

Hope this helps!

Great - makes sense. Thanks Jim!

Hey flupis,

Great question - and a pretty important one for understanding the concept of primes and factors.

All numbers (well, to be technical, all positive integers >1) are either prime or "composite" (the product of primes). So, looking at the numbers 2-10, we have:

2 = prime
3 = prime
4 = 2*2 (a composite number created by multiplying primes)
5 = prime
6 = 2*3 (composite)
7 = prime
8 = 2*2*2 (composite)
9 = 3*3 (composite)
10 = 2*5 (composite)

A number is only prime when its only factors are itself and 1. Well, 2^28 =

2*2*2..... (28 2s multiplied together). We know that this number will not be prime because it consists of multiple numbers multiplied together. And we know that 2 is the only prime factor because the only number we're multiplying in that set is 2 - we're just multiplying a ton of them. That's why 2 is the smallest prime factor of 2^28.

Makes sense thank you!