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Sequence S consists of 24 nonzero integers. If each term in

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Sequence S consists of 24 nonzero integers. If each term in [#permalink] New post 13 Nov 2006, 13:56
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A
B
C
D
E

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Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?

1) The third term in S is positive

2) The fourth term in S is negative
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 [#permalink] New post 13 Nov 2006, 14:22
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?

1) The third term in S is positive

2) The fourth term in S is negative

FROM ONE SEQUENCE COULD BE AS FOLLOWS

(+,+,+,+....) ALL TERMS ARE PSOITIVE OR

(-,-,+,-,-,+....) INSUFF

FROM TWO

THE SEQUENCE COULD BE

(-,+,-,-,+,-,-,+.....)

OR

(-,-,+,-,-,+,-,-,+,-....) ( SECONF SCENARIO)


INSUFF

BOTH TOGETHER

SECOND SCENARIO IS VALID .........SUFF

MY ANSWER IS C
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 [#permalink] New post 13 Nov 2006, 14:46
Is it B?

-,-,+,-,-,+
and
-,+,-,-,+,-,-,+ both have the same number of negative terms for 24 elements.
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 [#permalink] New post 13 Nov 2006, 17:27
I get C...

Set S has 24 elements ... how many are negative?

1) the thrid term is positive

lets see...

say 1st number is n1, second is n2...etc..

n1, n2, (n1*n2)=> positive, well if n1 and n2 are both negative then we have 1 number of negative elements..if n1 and n2 are both postive then we have 0 elements which are negative...INSUFF

2) fourth term is negative

n1, n2, [n1*n2], [n1*n2^2], this implies that n1 is negative..we dont know if n2 is negative or not..if it is..then we have a number of negative elements if not then we have another sufficient...

Combining them is sufficient..both n1 and n2 are negative...and we know exactly how many negative elements there are...
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 [#permalink] New post 13 Nov 2006, 19:54
I am also getting C.....
Exactly similar approach as of yezz, so don't want to type it...
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 [#permalink] New post 14 Nov 2006, 01:35
St1:
Not conclusive. We do not know if the 1st two terms are negative or positive.

St2:
If the 4th term is negative, then the product of the 2nd and 3rd term must yield negative.

If 3rd term is positive, then 2nd term must be negative and first term must negative. Total 16 negatives.

If 3rd term is negative, then 2nd term is positive and first term must be negative. Also 16 negatives.

So st2 is sufficient.

Ans B
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 [#permalink] New post 14 Nov 2006, 03:46
Very tricky stuff :shock: yup ans shld be B
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 [#permalink] New post 14 Nov 2006, 07:34
for me B as well
- - + - - + ......
- + - - + - ........
the number of negative integers is the same in both choices so B
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 [#permalink] New post 14 Nov 2006, 07:59
Is there are an easy way to find there are 16 - term in both the cases?
-,-,+,-.... and
-,+,-,- cases?
Or did you wrote down all the 24 terms and counted?
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 [#permalink] New post 14 Nov 2006, 09:07
It is a tricky question.

OA is B.

I was confused too. You only has to check all the possibilities for the first four numbers.

You will see that with the second condition it is enough.
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 [#permalink] New post 14 Nov 2006, 09:14
Nayan wrote:
Is there are an easy way to find there are 16 - term in both the cases?
-,-,+,-.... and
-,+,-,- cases?
Or did you wrote down all the 24 terms and counted?


Since it has a pattern, 2 -ves and 1+ve, I think we need not list everything, and can go with 2/3 of 24 for -ve. Not sure if there is any other way.
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 [#permalink] New post 14 Nov 2006, 09:23
There is another ways of solve this questions.

Nevertheless, I catch it when I wrote all the possibilities. I don't understand completely the other ways to solve it.

But the problem is that you spend more time.
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 [#permalink] New post 14 Nov 2006, 11:52
Hi Sumitra,
yes, there is a pattern -,-,+. But the first elements are differnt
--+
-+-
If you consider only 8 elements, in the first case, there are 6 -ivs, and 2 +ive.
In the 2nd sequence, there are 5-ivs and 3 +ivs.

So how do you know that 24 elements will have the same no. of negatives.[/quote]
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 [#permalink] New post 14 Nov 2006, 19:11
Nayan wrote:
Hi Sumitra,
yes, there is a pattern -,-,+. But the first elements are differnt
--+
-+-
If you consider only 8 elements, in the first case, there are 6 -ivs, and 2 +ive.
In the 2nd sequence, there are 5-ivs and 3 +ivs.

So how do you know that 24 elements will have the same no. of negatives.
[/quote]

Hi,
It is the same three --+/-+- that follows. Every three numbers should have 2 -ves and 1+ve, so out of 24, 16 -ves and 8 +ves.

However, I would say, when we know that it is gonna be the same --+/-+-
it would not take too long to list out the signs than to derive a shorter route.
  [#permalink] 14 Nov 2006, 19:11
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