To find the greatest prime factor of a number, we want to do a prime factorization - in other words, we want to break the number down as small as it'll go. But to do that, we first have to combine our terms.

When we see 4^17 - 2^28, we should notice right away that we can simplify 4 to 2^2. This will give us the same base for both of our terms (in this case, 2), so let's go ahead and do that. We'll need to remember our power rule: (x^m)^n = x^mn.

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

Now, we know that we can't add or subtract terms with the same base unless they *also* have the same exponent, so we'll have to find another way to simplify them. Thinking about the product rule ((x^m)(x^n) = x^(m+n)), we should recognize that 2^34 = 2^(28+6) = (2^28)(2^6). This means that we can pull 2^28 out of both terms:

(2^28)(2^6) - 2^28
(2^28)(2^6 - 1)

Now 2^28 has been factored as much as possible. We also have smaller numbers to subtract (2^6 and 1), so we can go ahead and combine them:

(2^28)(64 - 1)
(2^28)(63)

Then finally, we can do prime factorization on 63:

(2^28)(7)(9)
(2^28)(7)(3^2)

So the prime factors are 2, 3, and 7. Of these, the greatest prime factor is 7.

Fei wrote: ↑

Sat Mar 03, 2007 1:09 am

What is the greatest prime factor of:

4^17 - 2^28

Ans = 7... how to solve?

Tks+

We need to determine the greatest prime factor of 4^17 – 2^28. We can start by breaking 4^17 into prime factors.

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

Now our equation is as follows:

2^34 – 2^28

Note that the common factor in each term is 2^28; thus, the expression can be simplified as follows:

2^28(2^6 – 1)

2^28(64 – 1)

2^28(63)

2^28 x 9 x 7

2^28 x 3^2 x 7

We see that the greatest prime factor must be 7.