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

 It is currently 16 Oct 2018, 19:03

### 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

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

# If two integers are chosen at random out of the set {2, 5, 7, 8}, what

Author Message
TAGS:

### Hide Tags

Board of Directors
Joined: 17 Jul 2014
Posts: 2654
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

21 Feb 2016, 12:24
1
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
Manager
Joined: 24 May 2013
Posts: 79
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

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.
Intern
Joined: 17 Nov 2015
Posts: 8
Location: India
Concentration: Entrepreneurship, General Management
Schools: ISB '17
GMAT 1: 750 Q40 V35
GPA: 3.73
WE: Business Development (Energy and Utilities)
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

06 Apr 2016, 21:58
1
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
Intern
Joined: 05 Feb 2015
Posts: 3
If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

Updated on: 14 Mar 2018, 03:23
1
This is the best and easiest approach among all,

No's are of the form 4k+2 can not be written 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

Originally posted by suresh918 on 06 Jun 2016, 03:27.
Last edited by suresh918 on 14 Mar 2018, 03:23, edited 1 time in total.
Manager
Joined: 20 Mar 2015
Posts: 64
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, 8}, what  [#permalink]

### Show Tags

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!
Intern
Joined: 02 Sep 2016
Posts: 43
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

02 Oct 2016, 15:19
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
Intern
Joined: 07 Jun 2016
Posts: 39
GPA: 3.8
WE: Supply Chain Management (Manufacturing)
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

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
Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

28 Oct 2016, 04:32
1
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
Intern
Joined: 17 Aug 2016
Posts: 48
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8385
Location: Pune, India
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

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

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Joined: 17 Aug 2016
Posts: 48
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

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...
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8385
Location: Pune, India
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

10 Jan 2017, 03:50
1
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

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Senior Manager
Joined: 03 Apr 2013
Posts: 280
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

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.

=$$\frac{4}{6}$$

==$$\frac{2}{3}$$

(A)
_________________

Spread some love..Like = +1 Kudos

GMAT Tutor
Joined: 24 Jun 2008
Posts: 1327
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

13 Jul 2017, 08:00
1
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

Senior Manager
Joined: 28 Jun 2015
Posts: 293
Concentration: Finance
GPA: 3.5
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

24 Jul 2017, 22:08
2
My approach was similar to IanStewart's.

Possible products = $$(2*5), (2*7), (2*8), (5*7), (5*8), (7*8) = 10, 14, 16, 35, 40, 56$$.

In order for the product to be of the form $$(a^2 - b^2)$$, it has to be of the form $$(a+b)*(a-b)$$.

$$16 = (5+3) (5-3)$$
$$35 = (6+1) (6-1)$$
$$40 = (7-3) (7+3)$$
$$56 = (9+5) (9-5)$$

Probability = $$\frac{4}{6} = \frac{2}{3}$$. Ans - A.

Note: For the number to be of the form $$(a^2-b^2)$$ the difference between $$a$$ and $$b$$ must be an odd number.
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

Intern
Joined: 09 Jul 2017
Posts: 2
Location: India
Schools: Sloan '21
GMAT 1: 760 Q51 V41
GPA: 3.41
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

17 Aug 2017, 20:00
Can we simplify it as :

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

If (a-b) is odd, (a+b) has to be odd to give integral solutions and vice versa.

So (2,8), (5,7), (7,8), (5,8) all allow factoring into odd-odd or even-even pairs

Also by assigning a-b< a+b, we can get positive Integer solutions.

Since there 4C2 ways to select , 4/4C2 = 4/6 = 2/3
Intern
Joined: 14 Sep 2017
Posts: 11
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

30 Sep 2017, 09:48
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.

Ian, brilliant solution and thank you. One quick question. What about 8*8? 64 is the difference between 10^2 and 6^2. Does the equation capture this case? Are we assuming 2 distinct numbers?
Intern
Joined: 23 Sep 2017
Posts: 19
If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

Updated on: 13 Nov 2017, 03:48
I think here can be the easiest way to solve:
Sum of (a+b) and (a-b) always even => the product of two integers chosen in the set must factorize into 2 integers that are both even or odd.
We have: 2 straightforward visible couples that meet the requirement: 2 and 8; 5 and 7.
What else? Number 8 can factorize into 2 and 4. Therefore we can couple it with 5 and 7
=> We have 4 couple: (2,8), (5,7), (5,8), (7,8)
=> P = 4/ 8C2 = 2/3
Hope it helps

Originally posted by Designated Target on 11 Nov 2017, 14:10.
Last edited by Designated Target on 13 Nov 2017, 03:48, edited 1 time in total.
Director
Joined: 17 Dec 2012
Posts: 629
Location: India
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

12 Nov 2017, 01:32
The first step is to figure out that the product is of the form (a+b)(a-b) and that the two numbers that are picked are both even or both odd for a and b to be integers. We get two such pairs.

But the trick is a product can be formed by different pairs of numbers. So we need to look out for that too.

We have, 5*8 =40 written as (7+3)(7-3) and 7*8=56 written as (9+5)(9-5).

So now we can pick 4 pairs from the set satisfying the condition.

Total number of ways of picking 2 numbers is 4C2=6.

Probability is 4/6=2/3.
_________________

Srinivasan Vaidyaraman
Sravna Holistic Solutions
http://www.sravnatestprep.com

Holistic and Systematic Approach

Intern
Joined: 29 Aug 2017
Posts: 31
Location: India
GMAT 1: 640 Q49 V27
GPA: 4
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what  [#permalink]

### Show Tags

02 Jul 2018, 07:32
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.

I used long, time taking method to solve it.
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what &nbs [#permalink] 02 Jul 2018, 07:32

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

Display posts from previous: Sort by