First figure out the common factors in both

\(6^8\) can be written s \(2^8 * 3^8\)

\(take 3^8 common\)

\(3^8*2^8-3^8\)

\(3^8(2^8-1)\) (here you should know that \(2^8 = 256\), The easiest way to remember this is \(2^{10}=1024\) and now you can derive most \(2^x\))

SO your expression becomes \(3^8(256-1)\) ===> \(3^8 (255)\)==>\(3^8 (17*15)\)

so your prime factorisation will be \(3^8*15*17\)===>\(3^8*(3^1*5^1)*17\)=====>\(3^9*5^1*17\)

so as we can see there are there prime numbers here 3, 5 and 17 out of which 3 is the smallest prime number and 17 is the biggest.

Hence 17 is the biggest Prime Factor

Answer is C

zxcvbnmas wrote:

What is the greatest prime factor of \(6^8−3^8\) ?

A) 3

B) 11

C) 17

D) 19

E) 31

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