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:

45% (01:37) correct
55% (01:31) wrong based on 280 sessions

HideShow timer Statistics

If p and q are two different odd prime numbers, such that p < q, then which of the following must be true?

(A) (2p + q) is a prime number (B) p + q is divisible by 4 (C) q - p is divisible by 4 (D) (p + q + 1) is the difference between two perfect squares of integers (E) \(p^2 + q^2\) is the difference between two perfect squares of integers

Re: If p and q are two different odd prime numbers, such that [#permalink]

Show Tags

10 Jul 2013, 23:29

8

This post received KUDOS

1

This post was BOOKMARKED

mikemcgarry wrote:

If p and q are two different odd prime numbers, such that p < q, then which of the following must be true?

(A) (2p + q) is a prime number (B) p + q is divisible by 4 (C) q - p is divisible by 4 (D) (p + q + 1) is the difference between two perfect squares of integers (E) \(p^2 + q^2\) is the difference between two perfect squares of integers

Fact :Any odd number can be represented as \(2n\pm1\), where n is of-course an integer.

Scenario I. \(2n+1 = (n+1)^2-n^2 \to\)An odd number can be represented as difference of 2 perfect square

Scenario II. \(2n-1 = (n)^2-(n-1)^2 \to\) An odd number can be represented as difference of 2 perfect square

We know that p,q are both odd integers, thus p+q+1 = odd+odd+odd = 3*odd = odd. Thus, p+q+1 can always be represented as a difference of 2 perfect squares.

A. p=7,q=11, 2p+q = 25, not prime

B. p=7,q=11, p+q is not divisible by 4.

C.p=3,q=5,q-p is not divisible by 4.

D.Any odd number is ALWAYS a difference of 2 perfect square. Correct Answer.

E. For p=3,q=5, we have \(p^2+q^2 = 34\), Note that for the given problem, \(p^2+q^2\) will always be en even integer.

Now,any even integer can be represented as 2n, where n is an integer.

\((n+1)^2-(n-1)^2 = 4n = 2*2n\)Thus, any even number, which is a multiple of 4, can be represented as the difference of 2 perfect squares. As 34 is not divisible by 4, we can safely eliminate E.
_________________

Q. If p and q are two different odd prime numbers such that p < q, then which of the following must be true?

1.(2p + q) is a prime number 2. p + q is divisible by 4 3. q – p is divisible by 4 4. (p + q + 1) is the difference between two perfect squares of integers. 5. \((p^2+q^2)\) is the difference between two perfect squares of integers[/quote]

Analysis of the options : 1. (2p + q) for sure is an odd number , but may or may not be a prime number . Eg - if we take 3 and 5 as p and q resp, we get 11, which is a prime number. But if we take 7 and 11, we get 25, which is not a prime number. 2. p + q for sure gives us an even number. It may or may not be divisible by 4. Eg - (7 + 5) = 12 , which is divisible by 4. But (11 + 7) = 18 , not divisible by 4 3. Same way as 2nd. 4. (p + q + 1) gives us an odd number. Lets see the trend of difference between squares of consecutive integers. 2^2 - 1^2 = 3 3^2 - 2^2 = 5 4^2 - 3^2 = 7 5^2 - 4^2 = 9 6^2 - 5^2 = 11 and so on. The trend is : The difference between squares of consc. integers is always odd number. And it covers all the odd numbers.

However , when we don't take consecutive integers, this trend is not followed. Eg : 5^2 - 3^2 = 16 , which is an even number 3^2 - 1^2 = 8 , again not an odd number Hence : Any odd number can be expressed as the difference of the squares of integers.

5. (p^2 + q^2) gives an even number.
_________________

Quant Instructor ScoreBoost Bangalore, India

Last edited by samirchaudhary on 24 Mar 2014, 03:53, edited 2 times in total.

Your analysis still leaves a gap in my understanding of the problem as in explanation of answer choice D , you explain the difference between squares of consecutive integers part but then what about (p + q + 1). Same for 5 . Kindly explain and thanks.

[color=#898989]4. (p + q + 1) gives us an odd number. Lets see the trend of difference between squares of consecutive integers. 2^2 - 1^2 = 3 3^2 - 2^2 = 5 4^2 - 3^2 = 7 5^2 - 4^2 = 9 6^2 - 5^2 = 11 and so on. The trend is : The difference between squares of consc. integers is always odd number. However , when we don't take consecutive integers, this trend is not followed. Eg : 5^2 - 3^2 = 16 , which is an even number 3^2 - 1^2 = 8 , again not an even number Hence : for the option 'D' to be correct it has to be the difference between the squares of consc. integers. Please point out if I lost any point here. 5. (p^2 + q^2) gives an even number. Again doesn't make sense.[/quote][/color]
_________________

--------------------------

"The will to win, the desire to succeed, the urge to reach your full potential..."

Your analysis still leaves a gap in my understanding of the problem as in explanation of answer choice D , you explain the difference between squares of consecutive integers part but then what about (p + q + 1). Same for 5 . Kindly explain and thanks.

Dear royQV,

(p + q + 1) always gives us an odd number and the point I wanted to make from the analysis is that any odd number (read odd number we get by adding p,q and 1) can be expressed as the difference of the squares of consc integers.

Also for the 5th part, though we get even numbers as well as the difference of squares, all such even number may not be express-able as the difference of the squares of integers. <Not very clear though.>
_________________

Your analysis still leaves a gap in my understanding of the problem as in explanation of answer choice D , you explain the difference between squares of consecutive integers part but then what about (p + q + 1). Same for 5 . Kindly explain and thanks.

Dear royQV,

(p + q + 1) always gives us an odd number and the point I wanted to make from the analysis is that any odd number (read odd number we get by adding p,q and 1) can be expressed as the difference of the squares of consc integers.

Also for the 5th part, though we get even numbers as well as the difference of squares, all such even number may not be express-able as the difference of the squares of integers. <Not very clear though.>

Dear Samir & RoyQV, Yes, there are a few tricky number property shortcuts hidden in this problem. Appreciating these shortcuts could save time on a particularly challenging number property question on the test.

First of all, as Mr. Samir pointed out, any odd number can be expressed as the difference of squares of consecutive integers. You see, if we know (n^2), then all we have to do is add (n), then (n + 1), and that will result in (n + 1)^2.

For example, 7^2 = 49 49 + 7 + 8 = 64 = 8^2

20^2 = 400 400 + 20 + 21 = 441 = 21^2

This can be a very handy trick for finding squares close to square you already know (e.g. a multiple of ten). It also means that, given any odd number, we can easily express the odd number as a difference of squares. Any odd number can be written as the sum of two consecutive integers. For example, 71 = 35 + 36. Well, 71 must be the difference between those two squares, because (35^2) + 35 + 36 = (36^2) (35^2) + 71 = (36^2) Verify with a calculator that this pattern works for these numbers and for other numbers. That's the first pattern to know.

Now, the second pattern has to do with the difference of squares of two consecutive even numbers or two consecutive odd numbers. Those differences will always be multiples of 4. If you think about it algebraically, the difference between (n + 2)^2 = n^2 + 4n + 4 and n^2 is 4n + 4, which of course is divisible by four. Of course, all it takes is trying two odd prime numbers such as 3 and 5 to see that 3^2 + 5^2 = 9 + 25 = 34 is not something always divisible by 4, and therefore cannot be the difference between two odd or even consecutive numbers.

In fact, any odd number squared is a multiple of 4 plus 1, and any even number squares is a multiple of 4. Thus, the difference of any two odd squares or any two even squares would have to be a number divisible by 4. Again, it's good to check this with a calculator in hand until you verify for yourself that these patterns work.

Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: If p and q are two different odd prime numbers, such that [#permalink]

Show Tags

15 May 2014, 04:01

I concede with this theory. But there is a condition that those are consecutive numbers. If these numbers are not consecutive like 4^2 - 2^2 = 12, an even number. I think this statement is a contradictory one. Kindly correct me if I am wrong.

Re: If p and q are two different odd prime numbers, such that [#permalink]

Show Tags

15 May 2014, 05:37

1

This post received KUDOS

deya wrote:

I concede with this theory. But there is a condition that those are consecutive numbers. If these numbers are not consecutive like 4^2 - 2^2 = 12, an even number. I think this statement is a contradictory one. Kindly correct me if I am wrong.

Hi Deya, Your observation is right. When you have taken two even numbers, in which case difference of squares would be divisible by 4.

However, when you add two odd numbers, p and q and then add 1 i.e p+q+1,you will get an odd number. As shown before, any odd number is of the from 2k+1 Now, 2k+1 = k^2 + 2k + 1 - k^2 = (k+1)^2 - k^2 , which shows odd number = difference of two perfect squares (of two consecutive numbers)

Re: If p and q are two different odd prime numbers, such that [#permalink]

Show Tags

17 Aug 2015, 09:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If p and q are two different odd prime numbers, such that [#permalink]

Show Tags

28 Aug 2016, 15:51

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If p and q are two different odd prime numbers, such that [#permalink]

Show Tags

29 Aug 2016, 02:36

mikemcgarry wrote:

If p and q are two different odd prime numbers, such that p < q, then which of the following must be true?

(A) (2p + q) is a prime number (B) p + q is divisible by 4 (C) q - p is divisible by 4 (D) (p + q + 1) is the difference between two perfect squares of integers (E) \(p^2 + q^2\) is the difference between two perfect squares of integers