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If two integers are chosen at random out of the set {2, 5, 7 [#permalink]
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.
Re: From MitDavidDv: Choose Your Integers Question [#permalink]
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 **********************************
Re: From MitDavidDv: Choose Your Integers Question [#permalink]
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]
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.
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]
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]
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
Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.
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.
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]
27 Feb 2014, 22:13
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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 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.
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]
28 Feb 2014, 21:04
Expert's post
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 _________________
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
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. _________________
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]
05 Aug 2014, 12:31
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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
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]
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,
Re: If two integers are chosen at random out of the set {2, 5, 7 [#permalink]
15 Sep 2014, 01:21
Expert's post
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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,
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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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]
14 Jan 2015, 18:05
Very simple and straightforward method but never heard about the number property....
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
gmatclubot
Re: If two integers are chosen at random out of the set {2, 5, 7
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14 Jan 2015, 18:05
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