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Greatest prime factor...

Expert replies
Source: — Problem Solving |

by jayhawk2001 » Sat Mar 03, 2007 8:11 am
4^17 - 2^28
= 2^34 - 2^28
= 2^28 (2^6 - 1)
= 2^28 (63)
= 2^28 * 9 * 7

Prime factors are 2, 3 and 7. Greatest prime factor, hence, is 7
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re: greatest prime factor

by BTGmoderatorRO » Thu Aug 31, 2017 8:45 am
Mathematically, Prime factors originated from prime number. prime numbers which can be defined as any number that is divisible by itself and 1 alone. examples are 2, 3, 5, 7, 11 etc.
To every number, there are always factors which constitute the number. prime factors is finding which prime number we need to multiply to get the original numbers.

for the question, 4^17 - 2^28.
2^(2*17) - 2^28
2^34 - 2^28
2^28 (2^6 - 1)
2^28 * (64-1)
2^28 * (63)
2^28 * (7*9)
2^28 * (7 * 3^2)
2^28 * 3^2 * 7

for this solution, the prime factors are 2, 3 and 7. the greatest prime factor is the highest which of course is 7.
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by ErikaPrepScholar » Thu Aug 31, 2017 9:47 am
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.
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by Matt@VeritasPrep » Thu Aug 31, 2017 4:12 pm
More legibly :)

4¹� - 2²� =>

(2²)¹� - 2²� =>

2³� - 2²� =>

2²�*2� - 2²�*1 =>

2²� * (2� - 1) =>

2²� * 63 =>

2²� * 3 * 3 * 7
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Re: Greatest prime factor...

by Scott@TargetTestPrep » Fri Feb 14, 2020 1:35 pm
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.

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