Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

y1 , y2 , y3 .... are all distinct integers and represent [#permalink]

Show Tags

16 Mar 2007, 00:55

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

17% (00:00) correct
83% (01:46) wrong based on 16 sessions

HideShow timer Statistics

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

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 _________________

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. _________________

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

_________________

Want to improve your CR: http://gmatclub.com/forum/cr-methods-an-approach-to-find-the-best-answers-93146.html Tricky Quant problems: http://gmatclub.com/forum/50-tricky-questions-92834.html Important Grammer Fundamentals: http://gmatclub.com/forum/key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html

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

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. _________________

Want to improve your CR: http://gmatclub.com/forum/cr-methods-an-approach-to-find-the-best-answers-93146.html Tricky Quant problems: http://gmatclub.com/forum/50-tricky-questions-92834.html Important Grammer Fundamentals: http://gmatclub.com/forum/key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html

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. _________________

Want to improve your CR: http://gmatclub.com/forum/cr-methods-an-approach-to-find-the-best-answers-93146.html Tricky Quant problems: http://gmatclub.com/forum/50-tricky-questions-92834.html Important Grammer Fundamentals: http://gmatclub.com/forum/key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...