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If p and q are two different odd prime numbers, such that [#permalink]
10 Jul 2013, 10:08

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95% (hard)

Question Stats:

42% (02:32) correct
58% (01:32) wrong based on 173 sessions

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]
10 Jul 2013, 22:29

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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. _________________

Re: Odd Prime Numbers [#permalink]
22 Mar 2014, 23:04

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, 02:53, edited 2 times in total.

Re: Odd Prime Numbers [#permalink]
23 Mar 2014, 18:48

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] _________________

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Re: Odd Prime Numbers [#permalink]
24 Mar 2014, 03:00

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.> _________________

Re: Odd Prime Numbers [#permalink]
24 Mar 2014, 10:40

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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.

Re: If p and q are two different odd prime numbers, such that [#permalink]
14 May 2014, 13:11

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! shit-happens-my-journey-to-172475.html#p1372807

Re: If p and q are two different odd prime numbers, such that [#permalink]
15 May 2014, 03:01

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]
15 May 2014, 04:37

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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)

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Re: If p and q are two different odd prime numbers, such that
[#permalink]
15 May 2014, 04:37

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