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GMAT Prep- Problem [#permalink] New post 21 Mar 2006, 10:45
Can anyone help me with this problem? I ran across it in the test I was taking and can't figure out how to solve it.

4^17 – 2^28 (^ denotes to the power of)

What is the greatest prime factor?

Answer equals 7

Thanks in advance!
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 [#permalink] New post 21 Mar 2006, 10:49
4^17 – 2^28 = 2^34 - 2^28

2^28 (2^6 - 1)
2^28 (64 - 1)
(63)*(2^28)

63 = 7 * 3 * 3

So, the prime factors are 2, 3 and 7, and the greatest of these is 7
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Thanks [#permalink] New post 21 Mar 2006, 10:57
Thanks so much i was almost there but i was actually calculating it to equal 2^6 and getting 64.. thanks!!!
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 [#permalink] New post 21 Mar 2006, 12:25
jcgoodchild wrote:
4^17 – 2^28 = 2^34 - 2^28

2^28 (2^6 - 1)
2^28 (64 - 1)
(63)*(2^28)

63 = 7 * 3 * 3

So, the prime factors are 2, 3 and 7, and the greatest of these is 7


Jc please explain something to me

why is this wrong

2^34- 2^28 = 2^6

2^6 is 64

Closest prime to 64 is 63

63 = 7*3*3

where am I going wrong?
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 [#permalink] New post 21 Mar 2006, 12:36
X^a - X^b DOES NOT EQUAL X^(a-b)
however, X^a / X^b DOES EQUAL X^(a-b)


remember that adding exponants occur when you multipy and subtracting exponants occur only when you divide.

/MS
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 [#permalink] New post 21 Mar 2006, 12:46
Msam wrote:
X^a - X^b DOES NOT EQUAL X^(a-b)
however, X^a / X^b DOES EQUAL X^(a-b)


remember that adding exponants occur when you multipy and subtracting exponants occur only when you divide.

/MS


thanks you..
big mistake
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 [#permalink] New post 21 Mar 2006, 12:59
Where you went wrong is that 2^34- 2^28 does not equal 2^6. You can only do things to the exponents when you are multiplying or dividing.

So taking the originial equation 2^34 - 2^28, this is the same as saying 2^28*2^6 -- when you multiply numbers that have the same base but different exponents, you add the exponents; conversely, when you multiply numbers with same base but different exponents, you multiply the bases.. Does that make sense?
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 [#permalink] New post 21 Mar 2006, 13:07
jcgoodchild wrote:
Where you went wrong is that 2^34- 2^28 does not equal 2^6. You can only do things to the exponents when you are multiplying or dividing.

So taking the originial equation 2^34 - 2^28, this is the same as saying 2^28*2^6 -- when you multiply numbers that have the same base but different exponents, you add the exponents; conversely, when you multiply numbers with same base but different exponents, you multiply the bases.. Does that make sense?


yes that does make sense.

but where does the "-1 "come from in your answer
2^28*2^6-1.. or is that 2^28*2^6-2^28??

please explain.. that is still unclear to me
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 [#permalink] New post 21 Mar 2006, 13:39
Sorry, so let's break it down step-by-step

2^34 - 2^28

2^28*2^6 - 2^28*1

Now, since there is a common factor (2^28), we can pull that common factor out:

2^28 (2^6 - 1)

You can double check that this works by multiplying 2^28 through (2^6 - 1) to make sure that it is equal to the original equation.

Another explanation that might work would be if you had the following equation:

ax - a -- this simplifies to a(x - 1) since a is a common factor to both. In this specific quesiton, a = 2^28 and x = 2^6
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Got it [#permalink] New post 21 Mar 2006, 13:50
:idea: i knew there was somethign about the exponents that I was missing... that makes muuuch more sense!!! :o
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Re: Got it [#permalink] New post 21 Mar 2006, 14:08
AshikaP wrote:
:idea: i knew there was somethign about the exponents that I was missing... that makes muuuch more sense!!! :o


Yes I had that same feeling.

Thank you Jc... for that break down
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Re: Got it   [#permalink] 21 Mar 2006, 14:08
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