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prime factor - gmatprep

Expert replies
Source: — Problem Solving |

by pepeprepa » Sun Aug 10, 2008 6:28 am
4^17=2^34
So, 4^17-2^28=2^34-2^28=2^28(2^6-1)=2^28*63=2^28*3^2*7

You can see the biggest prime factor is 7 compared to 2 or 3
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by erjamit » Sun Aug 10, 2008 6:34 am
As a general strategy for exponent questions always try to reduce the terms either to lowest possible bases or to equal bases.

e.g. 4^4 - 16^2 ?

(This question I just made to illustrate the point ) :D

you can do two things...

1) 16^2 = (4^2)^2 = 4^4

2) 4^4 = (2^2)^4 = 2^8

16^2 = (2^4)^2 = 2^8

I hope it helps.

Thanks
Amit
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by smclean23 » Sun Aug 10, 2008 5:50 pm
From my simplification, I got 2^6.

Where does the factor of 7 come from?
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by umaa » Tue Sep 02, 2008 4:27 am
even i have the same question. I got 2^6. pls explain.
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by dferm » Tue Sep 02, 2008 5:33 am
Can someone please explain this question in a better format....?
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Re: prime factor - gmatprep

by sudhir3127 » Tue Sep 02, 2008 5:42 am
umaa wrote:i don't understand the answer. can you pls explain me.
here it is..

4^17 -2^28

we know we can write 4^17 as (2^2)17= 2^34

thus the whole thing now is

2^34- 2^28

take 2^28 as common

2^28( 2^6-1)

which is

2^28*63 ( 2^6= 64)

now prime factors of 63 are 3^2*7

thus we know that 7 is the highest prime factor.. hope its clear.
do let me know if u have any doubts..
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by dferm » Tue Sep 02, 2008 5:54 am
This is the part that is confusing me....

2^28( 2^6-1)
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by sudhir3127 » Tue Sep 02, 2008 6:14 am
dferm wrote:This is the part that is confusing me....

2^28( 2^6-1)
i will try to explain u.....

we know its

2^34-2^28

2^34 can also be written as 2^28*2^6 ( a^m*a*n = a^m+n )

therefore its now

2^28*2^6 - 2^28

Now we can take out 2^28 as common

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

15 - 5 can also be written as 5 ( 3-1)

hope it clears ur doubts ... do let me know if u still have doubts..
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by dferm » Tue Sep 02, 2008 7:18 am
GOT IT.....Ur the best... thanks for that lovely explanation... I was all over the place....
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