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# This is a problem from a review session that I do not

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Manager
Joined: 02 Jul 2003
Posts: 58

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This is a problem from a review session that I do not [#permalink]

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15 Aug 2003, 05:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

This is a problem from a review session that I do not attend.

If 2^n=128, then (2^n-1)(5^n-2)=

A. 10^7
B. 5(10^6)
C. 2(10^6)
D. 5(10^5)
E. 2(10^5)

answer: so I know n=7 then 2^5*5^5 is ? I multiply them out to get the answer E. Is there a faster solution where you get a common x so X^y*x^z= x^y+z
Thanks again,

Rich

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SVP
Joined: 03 Feb 2003
Posts: 1603

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15 Aug 2003, 05:58
n=7
(2^n-1)(5^n-2)= (2^6)(5^5)=2(2^5)(5^5)=2(2*5)^5=2*10^5

you should use brackets to make your formulae be more clear.

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15 Aug 2003, 05:58
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