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## Difficult Math Question #19 - Difficult Alegbra

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800guy Master | Next Rank: 500 Posts
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#### Difficult Math Question #19 - Difficult Alegbra

Fri Sep 22, 2006 10:03 am
OA coming after a few people answer. see image below..
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800guy Master | Next Rank: 500 Posts
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Tue Sep 26, 2006 5:53 pm
here's the OA:

The method I followed was to reduce the Q to (x^6/ x ) * ( Y/Y)
the eqn An= (x ^ n -1)(1+ x + x^2 + x^3 +.... ) ----------------------------(1)

and the eqn x(1+x(1+x(1+x...))))
which I call Z can be reduced to x( 1+x+x^2+x^3 ..) --------(2)

from (1) and (2) we get An / Z = x^(n-1) / x
therefore for getting answer x^5 (n-1) = 6
therefore n=7

Ans: B

gamemaster Senior | Next Rank: 100 Posts
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Fri Sep 22, 2006 1:05 pm
The expression in the multiple brackets equals eventually:

S = x+x^2+x^3+x^4+x^5

according to the question, An/S = x^5

=> An = x^6+x^7+x^8+x^9+x^10

so its easy to see n = 7

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