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

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If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 02 Jun 2011, 10:04
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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jul 2013, 12:17, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: From MitDavidDv: Choose Your Integers Question [#permalink] New post 03 Jun 2011, 11:49
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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 just used the exhaustive method. Count everything that fits.

2,5=10
2,7=14
2,8=16
5,7=35
5,8=40
7,8=56

Write down all perfect squares until 100
1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Pick one number at a time.
10 -- Keep adding with every perfect square and see whether the result is also there in the set.
10+1=11(Not there)
10+4=14(Not there)
10+9=19(Not there)
10+16=26(Not there)
10+25=35(Not there)

we can stop here as the difference between all consecutive perfect squares after 35 will be more than 10.
10- Not possible to represented as a^2-b^2
Repeat the same with all the products;
14-Not Possible

16: Let's check this;
16+1=17(Not there)
16+4=20(Not there)
16+9=25(There in the set)
16- can be represented as a^2-b^2 i.e. 5^2-3^2
**********************************

Likewise:
35- 6^2-1^2
40- 7^2-3^2
56- 9^2-5^2
*******************

In the sample set:
{10,14,16,35,40,56}
{10,14}- Not Possible: Count=2
{16,35,40,56}- Possible: Count=4

Total Count=6
P=Favorable/Total=4/6=2/3

Ans: "A"
**************************
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Re: From MitDavidDv: Choose Your Integers Question [#permalink] New post 03 Jun 2011, 17:50
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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.
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Re: From MitDavidDv: Choose Your Integers Question [#permalink] New post 19 Jul 2011, 23:06
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IanStewart
you gave a very easy explanation!
Thank you!
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Re: From MitDavidDv: Choose Your Integers Question [#permalink] New post 20 Jul 2011, 00:57
IanStewart,

Thanks for an easy explanation!!
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Re: From MitDavidDv: Choose Your Integers Question [#permalink] New post 23 Dec 2011, 05:31
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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.


Perfect. I didn't know that.
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 01 Oct 2013, 00:26
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 02 Oct 2013, 00:26
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Some Theory here:
Consider two numbers a, b
Now a* b always = [(a+b)/2]^ 2 – [(a-b)/2]^2---------------------- eqn 1
Additionally:
The number of ways a number can be expressed as a difference of two integers depends on number of ways it can be written as a two factor product stated below.
a.odd*odd
b. even *even
The reason being in eqn (1) above [(a-b)/2]^2 should result in an integer . Hence we consider only the above set of two factor products. (0dd minus odd = even ,even minus even =even , hence both will be divisible by 2

E.g the number 36 can be written as
6*6 , 18*2 hence 36 can be expressed as difference of squares in two ways.
36 = (6+6)/2 ^ 2 - (6-6)/2 ^ 2 = 6^2 - 0
36= (18+2)/2^2 – (18-2)/^2 = 10^2 – 4^2

Now back to the problem :
We have the number set (2, 5, 7, 8) out of which, the satisfying possibilities as per the above theory would be
2*8 (valid)
5*7 (valid)
2*5 (not valid)
2*7(not valid)
5*8== 4*10 hence valid
7*8 == 4*14 hence valid)
Hence there are 4 favourble cases out of 6. Therefore probability Is 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 28 Oct 2013, 20:20
Need Bunuel's explanation for this problem......
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 29 Oct 2013, 01:29
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 17 Nov 2013, 09:35
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Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem......


Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326

Bunuel please help out here..this will clear many things for me. My solution is this :-
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong.
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 27 Feb 2014, 22:13
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 have an explanation too, maybe it'd be of some help:
a^2 - b^2 =(a-b)(a+b)
(a+b) and (a-b) can only be integers from the selected set i.e {2,5,7,8}
Now a and b are both positive integers as stated in the question
So the sum of (a-b) and (a+b) also has to be an integer
i.e 2a = sum of any two numbers chosen from the set {2,5,7,8}

For a to be a positive integer the sum has to be an even number so either it could be a pair of (2,8) or (7,5)

Using PnC P(getting one such pair when chosing two random numbers from a set) = P(Both numbers chosen to be even) + P(Both numbers chosen to be odd)
= 2/4C2 + 2/4C2
= 4/4C2 = 4/6 = 2/3
Kindly let me know in case the solution has some mistakes.
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 28 Feb 2014, 21:04
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a^2-b^2 = (a+b)(a-b), so if the difference of any the factor pairs of the product is a positive even integer, then they can be in the a^2-b^2 form. Example: 7*5=35 --> 7*5 = (6+1)(6-1) --> 35*1 = (18-17)(18+17), etc etc

Only products that don't work are 2*5=10 (as 5-3 = odd and 10-1 = odd) and 2*7=14 (as 7-5 = odd and 14-1=13 odd).

The other 4 products work and so the probability = 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 08 Apr 2014, 03:43
ShashankDave wrote:
Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem......


Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326

Bunuel please help out here..this will clear many things for me. My solution is this :-
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong.



I had the same concern but now it's clear for me :
the 6 possible pairs are (2 5) (2 7) (2 8) (5 7) (5 8) (7 8)
our method allows to find 2 pairs (2 8) (5 7) but does not allow to eliminate the others
Especially (5 8 ) and (7 8) that also meet the condition : 5*8 = 10*4 = (7+3)*(7-3) and 7*8=14*4=(9+5)*(9-5)
so there are 4 possible pairs to be picked from 6 hence the probability is 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 13 Apr 2014, 19:09
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ShashankDave wrote:
Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem......


Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326

Bunuel please help out here..this will clear many things for me. My solution is this :-
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong.


The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers.

Two numbers are chosen and multiplied. Now they have lost their individual identity. Now you focus on the product and find whether it can be written as product of two numbers which are both odd or both even. Say you took two number 7 and 8 and multiplied them. You get 56. Can you write 56 as product of two numbers such that both are even? Yes, 4 and 14 or 2 and 28. So 56 can be written as a^2 - b^2 in two ways: (9^2 - 5^2) and (15^2 - 13^3). So if you choose 7, 8 from the set, their product can be written in the form a^2 - b^2.

Similarly, 5, 8 will give you the same result.

Hence you get 2 more cases and total probability becomes 4/6 = 2/3.

Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even.
If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number.
If the product is odd, it can always be written as product of two odd numbers.
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 05 Aug 2014, 12:31
Here is the solution with easy steps:

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

Now we need to find all possible combinations of two numbers from the set {2, 5, 7, 8 } which can be expressed as (a-b)(a+b)

Let say x =a-b and y = a+b, therefore x+y = 2a and y-x = 2b, so you need to have two numbers x and y whose sum and difference should be even number.

How many are there from the set {2, 5, 7, 8} ? 8 + 2 = 10 , 8 - 2 = 6, 7 - 5 = 2 , 7 + 5 = 12 . So there are two pairs (8,2) and (7,5) which can be expressed as a^2 - b^2.

Therefore, the probability that their product will be of the form a^2 – b^2 = 2/total two numbers combination = 2/4 Chose 2 = 2/3

Ans is option (A)
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 13 Sep 2014, 10:12
VeritasPrepKarishma wrote:

The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers.

Two numbers are chosen and multiplied. Now they have lost their individual identity. Now you focus on the product and find whether it can be written as product of two numbers which are both odd or both even. Say you took two number 7 and 8 and multiplied them. You get 56. Can you write 56 as product of two numbers such that both are even? Yes, 4 and 14 or 2 and 28. So 56 can be written as a^2 - b^2 in two ways: (9^2 - 5^2) and (15^2 - 13^3). So if you choose 7, 8 from the set, their product can be written in the form a^2 - b^2.

Similarly, 5, 8 will give you the same result.

Hence you get 2 more cases and total probability becomes 4/6 = 2/3.

Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even.
If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number.
If the product is odd, it can always be written as product of two odd numbers.

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,
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 15 Sep 2014, 01:21
Expert's post
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.
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Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink] New post 20 Oct 2014, 06:56
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
Re: If two integers are chosen at random out of the set {2, 5, 7   [#permalink] 20 Oct 2014, 06:56
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