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

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29 Mar 2016, 23:09

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.

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

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06 Apr 2016, 20:58

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

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

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06 Jun 2016, 02:27

1

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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

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

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21 Jul 2016, 09:03

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!

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

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02 Oct 2016, 14:19

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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

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

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03 Oct 2016, 16:17

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

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

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28 Oct 2016, 03:32

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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

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

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08 Jan 2017, 15:24

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,

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?

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,

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)
_________________

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...

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.
_________________

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

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12 Jul 2017, 23:34

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.

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.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]

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30 Sep 2017, 08:48

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?

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

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11 Nov 2017, 13:10

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

Last edited by Designated Target on 13 Nov 2017, 02:48, edited 1 time in total.

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.