It is currently 19 Sep 2017, 21:07

Close

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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If two integers are chosen at random out of the set {2, 5, 7

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

Hide Tags

Director
Director
User avatar
Joined: 07 Aug 2011
Posts: 584

Kudos [?]: 518 [0], given: 75

Concentration: International Business, Technology
GMAT 1: 630 Q49 V27
GMAT ToolKit User
If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 13 Mar 2015, 21:17
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6

Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.



So far if the average of the two numbers is an INTEGER they can be written in (a+b)(a-b) form . so that narrows us down to Odd + Odd and Even+Even cases .
Special consideration need to taken for those cases in which one number is ODD and other is multiple of 4 , i.e. in this case
if the set is \({ 2,5,7,8 }\), then possible pairs are :
7*8 = 56 = 14*4 = 28*2 none of these pairs (7,8) , (14,4), and (28*2) can be expressed in (a+b) (a-b) form .
5*8= 40 = 10*4 = (7+3) (7-3), so yes we can write \(5*8\) as \((7+3) * (7-3)\)
2,5,7,8
total number of cases = 4C2 = 6
favorable cases : (odd,odd) (5,7) , (Even,Even) (2,8) , and one special case as shown above (5,8) so \(3/6=1/2\)
_________________

Thanks,
Lucky

_______________________________________________________
Kindly press the Image to appreciate my post !! :-)

Kudos [?]: 518 [0], given: 75

Manhattan GMAT Discount CodesVeritas Prep GMAT Discount CodesJamboree Discount Codes
2 KUDOS received
Manager
Manager
User avatar
Joined: 10 Jun 2015
Posts: 126

Kudos [?]: 29 [2], given: 0

Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 11 Jun 2015, 22:19
2
This post received
KUDOS
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6

Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.

answer is (A)
the product set=(10, 14, 16, 35, 40, and 56)
16=8x2=(5+3)(5-3); (8-2)/2 =3
35=7x5=(6+1)(6-1); (7-5)/2 = 1
40=10x4=(7+3)(7-3)
56=14x4=(9+5)(9-5)
you got the pattern.

Kudos [?]: 29 [2], given: 0

1 KUDOS received
Intern
Intern
avatar
Joined: 05 Mar 2014
Posts: 6

Kudos [?]: 4 [1], given: 25

If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 05 Sep 2015, 20:45
1
This post received
KUDOS
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6

Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.


I consider

AxB = a^2 – b^2 = (a+b)(a-b)
where A and B are the posible chosen

first: the total posible chosen is 12, because A take 4 values and B takes 3, 4x3 = 12

Second
This is the scenary:

A x B = (a+b)(a-b)

2 5 = 10 = 2x5 or 10x1 (wrong) note that the sum of the factors should be even number, (conditions from a and b are integers), for the reason these opstion is eliminated
2 7 = 14 = 7x2 (wrong, the sum is not even) or 14x1 (wrong)
2 8 = 16 = 8x2 (correct)

5 2 = 10 = 8x2 (wrong, and the same that the first)
5 7 = 35 = 5x7 (Correct)
5 8 = 40 = 20x2 (correct)

7 2 = 14 = (wrong and is the same )
7 5 = 35 = 7x5 (correct)
7 8 = 56 = 14x4 (correct)

8 2 = 16 = 8x2 (correct)
8 5 = 40 = 20x2 (correct)
8 7 = 56 = 14x4 (correct)

Finally, the number of correct posible answer is 8, and the total possible answer is 12

indeed, 8/12 = 2/3

Kudos [?]: 4 [1], given: 25

1 KUDOS received
CEO
CEO
User avatar
G
Joined: 17 Jul 2014
Posts: 2589

Kudos [?]: 378 [1], given: 177

Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Premium Member Reviews Badge CAT Tests
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 21 Feb 2016, 12:24
1
This post received
KUDOS
only 2*8 and 7*5 will work.
the probability to choose each option is -> 1*1/3 = 1/3
since we have 2 options -> 1/3*2 = 2/3

Kudos [?]: 378 [1], given: 177

Manager
Manager
avatar
Joined: 24 May 2013
Posts: 86

Kudos [?]: 34 [0], given: 99

GMAT ToolKit User
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 30 Mar 2016, 00:09
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

So the general rule is that provided a number can be factorized with two even or two odd factors, then it can be represented as a difference of squares.
(5,7), (2,8), (5,8=10,4), (7,8 =4,14)=>4
Total cases =6
So 4/6=2/3 Hence A.

Kudos [?]: 34 [0], given: 99

Intern
Intern
User avatar
B
Joined: 17 Nov 2015
Posts: 9

Kudos [?]: [0], given: 0

Location: India
Concentration: Entrepreneurship, General Management
Schools: ISB '17
GMAT 1: 750 Q40 V35
GPA: 3.73
WE: Business Development (Energy and Utilities)
GMAT ToolKit User
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 06 Apr 2016, 21:58
Set = (2,5,7,8)
Let the product of two numbers to be selected from given set be x * y
Thus, x*y = a^2-b^2 = (a+b)*(a-b)
x = a + b
y = a - b
Solving these equations we get
x + y = 2a
x - y = 2b
therefore the two numbers we select must have summation & difference, an even number.
Amongst the 6 cases,
(2 , 5 ) -> difference is odd
(2 , 7) -> difference is odd
(2 , 8) -> summation & difference is even
(5 , 7) -> summation & difference is even
(5 , 8) -> this can be also expressed as 10 * 4 -> summation & difference is even
(7 , 8) -> this can be also expressed as 14 * 4 -> summation & difference is even

So out of 6 cases, 4 satisfy the equation
Therefore the required probablity = 4/6 = 2/3

Kudos [?]: [0], given: 0

Intern
Intern
avatar
Joined: 05 Feb 2015
Posts: 2

Kudos [?]: 1 [0], given: 2

If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 06 Jun 2016, 03:27
1
This post was
BOOKMARKED
This is the best and easiest approach among all,

No's are of the form 4k+2 can not write as a^2-b^2. for e.g. 6,10,.....etc. Remaining all no's we can write as a^2-b^2.
From the set,
2,5=10
2,7=14
2,8=16
5,7=35
5,8=40
7,8=56

Among all, 10,14 are 4k+2 no's so we can not write as a^2-b^2. un fav chances=2
remaining all we can write, so fav chances =4
probability =4/6=2/3

Kudos [?]: 1 [0], given: 2

Intern
Intern
avatar
Joined: 25 May 2016
Posts: 39

Kudos [?]: 6 [0], given: 136

Location: Singapore
Concentration: Finance, General Management
GPA: 2.8
WE: Engineering (Computer Software)
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 23 Jun 2016, 01:32
Is this a typical level 700 qn? For those who get a score above 700+ how many lvl 700 qns will they face?

Thank you Ian Stewart & Fluke! I like both your explanations.
_________________

Manickam

Kudos [?]: 6 [0], given: 136

Manager
Manager
User avatar
S
Joined: 20 Mar 2015
Posts: 70

Kudos [?]: 10 [0], given: 24

Location: United States
Concentration: General Management, Strategy
WE: Design (Manufacturing)
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 21 Jul 2016, 10:03
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6

Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.


Quickly looking at the products we have only 6 options (10,14,16,35,40 & 56) Now as a second step write squares from 1-10 (1,4,9,16,25,36,49,64,81,100) if we use unit digits to do a quick scan we can see (81-25, 49-9,36-1, 25-9) 4 cases are easily identified. The maximum probability in the options is 67% (2/3) so we don't have to search for more.
Ans: 4 out of 6 cases..(we combine both problem statement & options given to reach to the answer) Thanks!

Kudos [?]: 10 [0], given: 24

1 KUDOS received
Manager
Manager
User avatar
Joined: 30 Oct 2012
Posts: 68

Kudos [?]: 21 [1], given: 19

Location: India
WE: Marketing (Manufacturing)
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 29 Aug 2016, 00:29
1
This post received
KUDOS
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6

Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.


good one.

thanks to Ian Stewart for his explanation.
_________________

It’s not over until I say it’s over

Kudos [?]: 21 [1], given: 19

Intern
Intern
User avatar
Joined: 02 Sep 2016
Posts: 47

Kudos [?]: 40 [0], given: 130

Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 02 Oct 2016, 15:19
1
This post was
BOOKMARKED
bsmith37 wrote:
not sure if its been discussed, but a valuable property to know is that ANY non-prime odd number, or multiple of 4, can be written as a difference of squares using integers. 21 = (5+2)(5-2) 15 = (4+1)(4-1) etc. try it out.

therefore, we can see that out of our 6 possible outcomes, only 4 will be either odd (5 x 7) or multiples of 4 (8 x each other #). so answer = 4/6=2/3


This solution is the best amongst all...time saving and straightforward 8-)

Kudos [?]: 40 [0], given: 130

Intern
Intern
avatar
Joined: 07 Jun 2016
Posts: 47

Kudos [?]: 7 [0], given: 106

GPA: 3.8
WE: Supply Chain Management (Manufacturing)
GMAT ToolKit User Premium Member CAT Tests
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 03 Oct 2016, 17:17
IanStewart wrote:
You can avoid an exhaustive test here. Suppose I ask whether (97)(103) can be written in the form a^2 - b^2, where a and b are integers. Notice that this is a difference of squares: a^2 - b^2 = (a+b)(a-b). We can now just use the median of 97 and 103, which is 100:

(97)(103) = (100-3)(100+3) = 100^2 - 3^2

So whenever we can write our product in such a way that the median of our two numbers is an integer, we can write our product as a difference of squares just as above. For example, if we take 5*7, that's equal to (6-1)(6+1), and if we take 2*8, that's equal to (5-3)(5+3). Now if we look at 8*5, we can't immediately use the same trick, but we can 'move' one of the 2s from the 8 into the 5, as follows: 8*5 = 4*10 = (7-3)(7+3). Similarly, 8*7 = 4*14 = (9-5)(9+5). So of our six possible products, four can be written as a difference of squares.


that was an excellent explanation, thank you very much Ian. I actually did it the way the first person did but it was definitely exhaustive! I am going to practice this in case it shows up on the test

Kudos [?]: 7 [0], given: 106

Senior Manager
Senior Manager
avatar
B
Joined: 13 Oct 2016
Posts: 367

Kudos [?]: 387 [0], given: 40

GPA: 3.98
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 28 Oct 2016, 04:32
1
This post was
BOOKMARKED
The number can be written as difference of 2 squares only in two cases: when it’s odd or in case when it’s even - it should be a multiple of 4 (even number can be represented as 2x, putting this to the formulae of difference of squares we’ll have common factor of 4). Now let’s make different cases depending on the first number chosen.
If the first number was chosen to be 2 (probability ¼). The only number, that will satisfy the given condition will be 8 (probability 1/3) Resulting product (1/4)*(1/3)=1/12
Next cases will be following:
First number chosen - 5(prob ¼), numbers that satisfy given condition – 7,8 (prob 2/3) result (1/4)*(2/3)=2/12
First number chosen - 7(prob ¼), numbers that satisfy given condition – 5,8 (prob 2/3) result (1/4)*(2/3)=2/12
First number chosen - 8(prob ¼), numbers that satisfy given condition –2, 7,5 (prob 1= 3/3) result (1/4)*(3/3)=3/12
Cases are independent so summing up all partial results we’ll get :
1/12+2/12+2/12+3/12=8/12=2/3
Answer A

Kudos [?]: 387 [0], given: 40

Manager
Manager
avatar
B
Joined: 17 Aug 2016
Posts: 50

Kudos [?]: 1 [0], given: 82

Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 08 Jan 2017, 16:24
VeritasPrepKarishma wrote:
ronr34 wrote:
Hi Karishma,
Can you elaborate a little more?
Why are we looking for 2 numbers that are either both even or both odd?
Also, how did you know to stop at (9^2 - 5^2) and (15^2 - 13^3) and not look for more?
Thanks,


That's a good question. You should understand this concept well. That is why I have written a detailed post on it on my blog:
http://www.veritasprep.com/blog/2014/04 ... at-part-i/

Check it out and get back to me (on the blog or here) if any doubts remain.


Hi Karishma, your blog is awesome, as it is your explanation to this problem.

My only doubt is with regards to the number 4. You said it is sufficient that we have the number for to be able to write the product of 2 integers in the form a^2-b^2.
But this doesn't seem to work if I select 1 and 4. It looks to me this is the only case it doesn't work. Am I right?

Thanks

Kudos [?]: 1 [0], given: 82

Expert Post
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7604

Kudos [?]: 16876 [0], given: 230

Location: Pune, India
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 09 Jan 2017, 08:25
bazu wrote:
VeritasPrepKarishma wrote:
ronr34 wrote:
Hi Karishma,
Can you elaborate a little more?
Why are we looking for 2 numbers that are either both even or both odd?
Also, how did you know to stop at (9^2 - 5^2) and (15^2 - 13^3) and not look for more?
Thanks,


That's a good question. You should understand this concept well. That is why I have written a detailed post on it on my blog:
http://www.veritasprep.com/blog/2014/04 ... at-part-i/

Check it out and get back to me (on the blog or here) if any doubts remain.


Hi Karishma, your blog is awesome, as it is your explanation to this problem.

My only doubt is with regards to the number 4. You said it is sufficient that we have the number for to be able to write the product of 2 integers in the form a^2-b^2.
But this doesn't seem to work if I select 1 and 4. It looks to me this is the only case it doesn't work. Am I right?

Thanks


We need 4 in the product to be able to write the product as the product of two even numbers.

Say the product is 4. You can write it as 2*2 (both even)
Say the product is 4*3 = 12. You can write it as 2*6 (both even)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 16876 [0], given: 230

Manager
Manager
avatar
B
Joined: 17 Aug 2016
Posts: 50

Kudos [?]: 1 [0], given: 82

If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 09 Jan 2017, 16:58
VeritasPrepKarishma wrote:
bazu wrote:
VeritasPrepKarishma wrote:

That's a good question. You should understand this concept well. That is why I have written a detailed post on it on my blog:
http://www.veritasprep.com/blog/2014/04 ... at-part-i/

Check it out and get back to me (on the blog or here) if any doubts remain.


Hi Karishma, your blog is awesome, as it is your explanation to this problem.

My only doubt is with regards to the number 4. You said it is sufficient that we have the number for to be able to write the product of 2 integers in the form a^2-b^2.
But this doesn't seem to work if I select 1 and 4. It looks to me this is the only case it doesn't work. Am I right?

Thanks


We need 4 in the product to be able to write the product as the product of two even numbers.

Say the product is 4. You can write it as 2*2 (both even)
Say the product is 4*3 = 12. You can write it as 2*6 (both even)



But the in then case of 2*2, how can i write the product in the form a^2-b^2? Apologies for the silly question, but I can't find an answer...

Kudos [?]: 1 [0], given: 82

Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7604

Kudos [?]: 16876 [1], given: 230

Location: Pune, India
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 10 Jan 2017, 03:50
1
This post received
KUDOS
Expert's post
bazu wrote:

But the in then case of 2*2, how can i write the product in the form a^2-b^2? Apologies for the silly question, but I can't find an answer...


\(2*2 = (2+0)*(2 - 0) = 2^2 - 0^2\)

Of course, here we need both a and b to be positive integers so it won't work. Anyway, we need to pick 2 of the given 4 numbers. Their product will be more than 4.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 16876 [1], given: 230

Manager
Manager
avatar
B
Joined: 17 Aug 2016
Posts: 50

Kudos [?]: 1 [0], given: 82

Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 11 Jan 2017, 04:51
VeritasPrepKarishma wrote:
bazu wrote:

But the in then case of 2*2, how can i write the product in the form a^2-b^2? Apologies for the silly question, but I can't find an answer...


\(2*2 = (2+0)*(2 - 0) = 2^2 - 0^2\)

Of course, here we need both a and b to be positive integers so it won't work. Anyway, we need to pick 2 of the given 4 numbers. Their product will be more than 4.



Brilliant, thank you very much!

Kudos [?]: 1 [0], given: 82

Manager
Manager
User avatar
S
Joined: 03 Apr 2013
Posts: 241

Kudos [?]: 36 [0], given: 823

GMAT ToolKit User
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 13 Jul 2017, 00:34
IanStewart wrote:
You can avoid an exhaustive test here. Suppose I ask whether (97)(103) can be written in the form a^2 - b^2, where a and b are integers. Notice that this is a difference of squares: a^2 - b^2 = (a+b)(a-b). We can now just use the median of 97 and 103, which is 100:

(97)(103) = (100-3)(100+3) = 100^2 - 3^2

So whenever we can write our product in such a way that the median of our two numbers is an integer, we can write our product as a difference of squares just as above. For example, if we take 5*7, that's equal to (6-1)(6+1), and if we take 2*8, that's equal to (5-3)(5+3). Now if we look at 8*5, we can't immediately use the same trick, but we can 'move' one of the 2s from the 8 into the 5, as follows: 8*5 = 4*10 = (7-3)(7+3). Similarly, 8*7 = 4*14 = (9-5)(9+5). So of our six possible products, four can be written as a difference of squares.


IanStewart As an extension to your method, this is how I did it. Please see if my reasoning is correct.

First of all

\(a^2-b^2 = (a+b)(a-b)\)

Here, both a and b are positive integers and so will be their sum or product.

For any product that can be rewritten in this manner, there has to be two integers (a+b) and (a-b) whose product the number will be.

Now, what happens if we add these two integers?

\((a+b) + (a-b) = 2a\)

Thus, For it to be possible for a number to rewritten in this form, it must be possible for the number to be written as a product of any two numbers whose sum is an even integer. In other words, the two multiplied numbers must be either both even or both odd. Taking examples.


2,5
2*5
There is no other method to write this product, and since the sum is 7, this doesn't count.

2,7
2*7
sum 9, doesn't count

2,8
2*8
sum 10, yes this counts.

5,7
5*7
sum 12, yes.

5,8
5*8
another way
10*4
sum 14, yes.

see where I'm getting at?

7,8
7*8
another way
14*4
sum 18, yes.

This out of 6 possible selections, 4 count as favorable.

Final answer

=\(\frac{4}{6}\)

==\(\frac{2}{3}\)

(A)
_________________

Spread some love..Like = +1 Kudos :)

Kudos [?]: 36 [0], given: 823

Expert Post
1 KUDOS received
GMAT Tutor
avatar
B
Joined: 24 Jun 2008
Posts: 1341

Kudos [?]: 1908 [1], given: 6

If two integers are chosen at random out of the set {2, 5, 7 [#permalink]

Show Tags

New post 13 Jul 2017, 08:00
1
This post received
KUDOS
Expert's post
ShashankDave wrote:

Now, what happens if we add these two integers?

\((a+b) + (a-b) = 2a\)

Thus, For it to be possible for a number to rewritten in this form, it must be possible for the number to be written as a product of any two numbers whose sum is an even integer. In other words, the two multiplied numbers must be either both even or both odd. Taking examples.

5,8
5*8
another way
10*4
sum 14, yes.



You're definitely on the right track if you want to prove exactly when it's possible to do this, but when you take examples, unless I've misunderstood what you're saying, I think you're assuming the numbers in the product are equal to a and b. But if you look, say, at this product:

10*4 = (7 + 3)(7 - 3)

then the values of a and b are 7 and 3, respectively, and not 10 and 4. That's why the conclusion you've written, "the two multiplied numbers must be either both even or both odd" is technically not right, as your example of 5*8 demonstrates (there we have one odd and one even, but we can still write the product as a difference of squares). Your conclusion is very close to being right, though - it's right if you rephrase it: "it must be possible to somehow write the product as a product of two even or two odd numbers".

You could modify your proof just slightly, and get a cleaner statement of when it is possible to do this. As you correctly point out, the sum of a+b and a-b must be even. So the two numbers a+b and a-b must either both be odd, or both be even. Thus the product (a+b)(a-b) is either odd, or it is the product of two even numbers, and is thus divisible by 4. And that's it: those are the two circumstances where it's possible to write a product of two integers as a difference of squares. Either our product must be odd, or it must be a multiple of 4. So we can do this in any of these situations:

product of two odds: e.g. 5*17 = (11 - 6)(11 + 6)
product of two evens: e.g. 14*24 = (19 - 5)(19 + 5)
product of one odd and one multiple of 4 (by moving one 2 from the even number to the odd number, to get two even numbers) : e.g. 11*16 = 22*8 = (15 + 7)(15 - 7)

and we can always do it in those situations, because when we have a product of two odds or of two evens, the median of our two numbers will always be an integer, and then we can always use the 'trick' above.

But we cannot do it in this situation:

product of one even number that is *not* divisible by 4 and one odd number: 14*5

because a^2 - b^2 must always be either odd or divisible by 4. There are a few other ways to prove that - for example, when a and b are both even, a^2 and b^2 are both divisible by 4, so a^2 - b^2 must be divisible by 4, and when a and b are both odd, you can just plug a = 2s + 1 and b = 2t + 1 into a^2 - b^2, and you'll see in that case a^2 - b^2 is also divisible by 4. So whenever a^2 - b^2 is even, it is always divisible by 4, which is what we established in a different way above.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Kudos [?]: 1908 [1], given: 6

If two integers are chosen at random out of the set {2, 5, 7   [#permalink] 13 Jul 2017, 08:00

Go to page   Previous    1   2   3    Next  [ 42 posts ] 

    Similar topics Author Replies Last post
Similar
Topics:
11 EXPERTS_POSTS_IN_THIS_TOPIC Two integers x and y are chosen without replacement out of the set {1, Bunuel 12 06 May 2017, 05:40
5 EXPERTS_POSTS_IN_THIS_TOPIC If two integers between –5 and 3, inclusive, are chosen at random, whi gmat1220 5 06 May 2017, 05:50
1 If a number N is chosen at random from the set of two-digit integers gmat1220 4 23 Apr 2011, 12:43
1 EXPERTS_POSTS_IN_THIS_TOPIC n is an integer chosen at random from the set {5, 7, 9, 11 } p is Bunuel 3 19 May 2016, 07:23
8 EXPERTS_POSTS_IN_THIS_TOPIC if set a ={2,3,5,7} and two numbers are selected at random snorkeler 5 16 Jun 2016, 13:02
Display posts from previous: Sort by

If two integers are chosen at random out of the set {2, 5, 7

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.