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

Question

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

Fig · Answer

Direct (E)

All values of x provide P = 0

.... There is no maximum or minimum but only equaled values of P to 0

Amardeep Sharma · Answer

I agree... infact for any integer its true, whenever Yn=X...

regards,

Amardeep

GMAThopeful · Answer

Yup.. E is the answer.. I got it wrong though

bibha · Answer

Hey, I was going through this problem. Somehow, i could not understand the explanation provided. Please, help me through this. Bibha :-)

hardnstrong · Answer

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

Silvers · Answer

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.

ykaiim · Answer

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 · Answer

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

ykaiim · Answer

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.

nsp007 · Answer

great explanations !

ykaiim · Answer

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.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards