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# ds og 135 geometric sequence ?

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Manager
Joined: 05 Oct 2005
Posts: 200

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ds og 135 geometric sequence ? [#permalink]

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23 Nov 2005, 14:52
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a good one :

if s is the infinite sequence S^1 ( first term)=9
S^2=99, S^3=999.....

S^k=10^k-1...

IS every term in S divisible by the prime number p?

1 p is greater than 2

2 AT least one term in the sequence is divisible by p

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Manager
Joined: 05 Oct 2005
Posts: 200

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23 Nov 2005, 15:11
nakib77 wrote:
It should be B.

nope
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VP
Joined: 06 Jun 2004
Posts: 1051

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Location: CA
Re: ds og 135 geometric sequence ? [#permalink]

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23 Nov 2005, 15:28

(1) p is greater than 2 so p could be 3, 5, 7.....Insufficient
(2) p could be 11...so s^2 will be divisible by 11, but s^3 will not. Insufficient.

(1) + (2) still not sufficient based on above as p could be 3 and 11...

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Director
Joined: 09 Jul 2005
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23 Nov 2005, 16:26
E for me

(1) p=3 and p=5 then NS

(2) 11 divides 99 but it doesn't 999. 3 does for both. NS

(1) and (2). 11 divides 99 but it doesn't 999. 3 does for both. NS

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Manager
Joined: 05 Oct 2005
Posts: 200

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25 Nov 2005, 16:45
OA IS E
THANKS
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when there is a will there is a way

best regards

Kudos [?]: 7 [0], given: 0

25 Nov 2005, 16:45
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