GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 May 2019, 16:07 ### GMAT Club Daily Prep

#### 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. ### Request Expert Reply # If p and q are two different odd prime numbers, such that

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4485
If p and q are two different odd prime numbers, such that  [#permalink]

### Show Tags

5
16 00:00

Difficulty:   95% (hard)

Question Stats: 46% (02:38) correct 54% (02:26) wrong based on 356 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

For a discussion of "must be true" problems, as well as a complete solution to this question, see:
http://magoosh.com/gmat/2013/gmat-quant ... -problems/

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)
##### Most Helpful Expert Reply
Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4485
Re: Odd Prime Numbers  [#permalink]

### Show Tags

7
6
samirchaudhary wrote:
Dear Samir,

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)
##### Most Helpful Community Reply
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 611
Re: If p and q are two different odd prime numbers, such that  [#permalink]

### Show Tags

8
2
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

For a discussion of "must be true" problems, as well as a complete solution to this question, see:
http://magoosh.com/gmat/2013/gmat-quant ... -problems/

Mike 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.
_________________
##### General Discussion
Intern  Affiliations: ScoreBoost
Joined: 07 Sep 2012
Posts: 14
Location: India
Re: Odd Prime Numbers  [#permalink]

### Show Tags

1
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

Originally posted by samirchaudhary on 23 Mar 2014, 00:04.
Last edited by samirchaudhary on 24 Mar 2014, 03:53, edited 2 times in total.
Manager  Joined: 24 Apr 2012
Posts: 50
Concentration: Strategy
WE: Project Management (Consulting)
Re: Odd Prime Numbers  [#permalink]

### Show Tags

Dear Samir,

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..."

Useful Links:

http://gmatclub.com/forum/powerscore-cr-notes-hope-that-someone-find-it-kudosofiable-174638.html#p1384561
Intern  Affiliations: ScoreBoost
Joined: 07 Sep 2012
Posts: 14
Location: India
Re: Odd Prime Numbers  [#permalink]

### Show Tags

Dear Samir,

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.>
_________________
Quant Instructor
ScoreBoost
Bangalore, India
Intern  Joined: 12 Sep 2012
Posts: 25
GMAT 1: 550 Q49 V17 Re: If p and q are two different odd prime numbers, such that  [#permalink]

### Show Tags

Kindly explain why it is (D). It is not very clear to me.
Senior Manager  Joined: 28 Apr 2012
Posts: 275
Location: India
Concentration: Finance, Technology
GMAT 1: 650 Q48 V31 GMAT 2: 770 Q50 V47 WE: Information Technology (Computer Software)
Re: If p and q are two different odd prime numbers, such that  [#permalink]

### Show Tags

1
deya wrote:
Kindly explain why it is (D). It is not very clear to me.

read explanation by maul.
Quote:
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

Since p and q both are odd, p+q+1 will be odd. And any odd number can be represented as difference between two perfect squares.
_________________
"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well."
― Voltaire

Press Kudos, if I have helped.
Thanks!
Intern  Joined: 12 Sep 2012
Posts: 25
GMAT 1: 550 Q49 V17 Re: If p and q are two different odd prime numbers, such that  [#permalink]

### Show Tags

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.
Intern  Joined: 13 May 2014
Posts: 33
Concentration: General Management, Strategy
Re: If p and q are two different odd prime numbers, such that  [#permalink]

### Show Tags

1
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)

Best form of appreciation is kudos
Manager  B
Joined: 30 Oct 2012
Posts: 62
Location: India
WE: Marketing (Manufacturing)
Re: If p and q are two different odd prime numbers, such that  [#permalink]

### Show Tags

1
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

For a discussion of "must be true" problems, as well as a complete solution to this question, see:
http://magoosh.com/gmat/2013/gmat-quant ... -problems/

Mike p+q+1 will be odd.
difference of two perfect square numbers is also odd
_________________
i am the master of my fate, I am the captain of my soul
Non-Human User Joined: 09 Sep 2013
Posts: 11007
Re: If p and q are two different odd prime numbers, such that  [#permalink]

### Show Tags

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] 18 Dec 2018, 20:13
Display posts from previous: Sort by

# If p and q are two different odd prime numbers, such that

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

#### MBA Resources  