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y1 , y2 , y3 .... are all distinct integers and represent

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y1 , y2 , y3 .... are all distinct integers and represent [#permalink] New post 15 Mar 2007, 23:55
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y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers. P is a non negative number such that
P = (x - y1 ) (x - y2 ) (x - y3 ) .... upto infinity, where x is an integer.

Which of the following value can x take so that p has the least value?
I. 0 II. 1 III. -1


Answer Choices :

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) any of I, II or III
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 [#permalink] New post 16 Mar 2007, 13:53
Direct (E) :)

All values of x provide P = 0 :).... There is no maximum or minimum but only equaled values of P to 0 :)
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 [#permalink] New post 16 Mar 2007, 22:20
I agree... infact for any integer its true, whenever Yn=X...

regards,

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[#permalink] New post 16 Mar 2007, 23:52
Yup.. E is the answer..
I got it wrong though :(
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Re: PS [#permalink] New post 04 May 2010, 19:36
Hey,
I was going through this problem. Somehow, i could not understand the explanation provided. Please, help me through this.
Bibha :-)
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Re: PS [#permalink] New post 05 May 2010, 03:07
GMAThopeful wrote:
y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers. P is a non negative number such that
P = (x - y1 ) (x - y2 ) (x - y3 ) .... upto infinity, where x is an integer.

Which of the following value can x take so that p has the least value?
I. 0 II. 1 III. -1


Answer Choices :

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) any of I, II or III



IMO E
here's my explanation-

questions say "y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers"
so one is a negative integer, one is positive and third zero
Now the least value of P will be zero because it goes upto infinity. Put in the values of y1, y2, y3 to make p=0


All three satisfies the condition.
Correct me if i am wrong
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Re: PS [#permalink] New post 05 May 2010, 10:03
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GMAThopeful wrote:
y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers. P is a non negative number such that
P = (x - y1 ) (x - y2 ) (x - y3 ) .... upto infinity, where x is an integer.

Which of the following value can x take so that p has the least value?
I. 0 II. 1 III. -1


Answer Choices :

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) any of I, II or III



Its E.

since y1,y2 ...yn are disticnt integers they will be equal to 0,-1 and 1 at some point or the other and at that point

x = 0 will give (0-0) so P=0
x=1 will be 0 when y = -1
x =-1 will be 0 at y = 1

so no matter what value of x we take the value of P will be 0.
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Re: PS [#permalink] New post 05 May 2010, 10:31
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It's E but it is not mandatory that the value of P = 0 in all the cases.

Let's see how: say y1, y2, y3,... = -4, -2, 0, 2, 4

I. x=0 >>> P = (0+4)(0+2)(0+0)(0-2)(0-4)... = 0
II. 1 >>> P = (1+4)(1+2)(1+0)(1-2)(1-4)... = Non-zero
III. -1 >>> P = (-1+4)(-1+2)(-1+0)(-1-2)(-1-4)... = Non-zero

Hope this helps.

hardnstrong wrote:
GMAThopeful wrote:
y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers. P is a non negative number such that
P = (x - y1 ) (x - y2 ) (x - y3 ) .... upto infinity, where x is an integer.

Which of the following value can x take so that p has the least value?
I. 0 II. 1 III. -1


Answer Choices :

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) any of I, II or III



IMO E
here's my explanation-

questions say "y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers"
so one is a negative integer, one is positive and third zero
Now the least value of P will be zero because it goes upto infinity. Put in the values of y1, y2, y3 to make p=0


All three satisfies the condition.
Correct me if i am wrong

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Re: PS [#permalink] New post 07 May 2010, 04:44
ykaiim wrote:
It's E but it is not mandatory that the value of P = 0 in all the cases.

Let's see how: say y1, y2, y3,... = -4, -2, 0, 2, 4

I. x=0 >>> P = (0+4)(0+2)(0+0)(0-2)(0-4)... = 0
II. 1 >>> P = (1+4)(1+2)(1+0)(1-2)(1-4)... = Non-zero
III. -1 >>> P = (-1+4)(-1+2)(-1+0)(-1-2)(-1-4)... = Non-zero

Hope this helps.

hardnstrong wrote:
GMAThopeful wrote:
y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers. P is a non negative number such that
P = (x - y1 ) (x - y2 ) (x - y3 ) .... upto infinity, where x is an integer.

Which of the following value can x take so that p has the least value?
I. 0 II. 1 III. -1


Answer Choices :

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) any of I, II or III



IMO E
here's my explanation-

questions say "y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers"
so one is a negative integer, one is positive and third zero
Now the least value of P will be zero because it goes upto infinity. Put in the values of y1, y2, y3 to make p=0


All three satisfies the condition.
Correct me if i am wrong



Question is asking for the least value of P , which can be 0 only(which is possible in this case) as P is a non negative number
if you find value of P as zero in one case and non zero in otherwise. that way you contradicts the correct answer that all three values satifies. Here we have to prove if all three conditions could satisfy it or not. Not necessarily true in all cases
You example shows that only one value satisfies. Hence changes the answer

Hope this helps
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Re: PS [#permalink] New post 07 May 2010, 05:45
Agree. To become the least value, P is to be 0 or -ve.

I just took a sample to prove that it is not neccessary to have P = 0 as the only option. Since, the series is infinitly long so you can take as many values in the series.
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Re: PS [#permalink] New post 08 May 2010, 22:34
great explanations !
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Re: PS [#permalink] New post 08 May 2010, 23:56
A few more things. It is not mentioned that:

1. Order is not mentioned that y1, y2, y3... can be concesutive numbers. So, this series can be >>> (-2, 0, 2, 4...), (-2, -1, 0, 1, 2...), (-5, -4, 0, 2, 10...)
2. P is non-negative, which means it can be 0 or a +ve number and moreover a limit on value of P is not mentioned.

So, my cases are correct.
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Re: PS   [#permalink] 08 May 2010, 23:56
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