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

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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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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.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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 am sorry, i still dont know how you got the answer from this. What to do after these steps.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
lostaish wrote:
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 am sorry, i still dont know how you got the answer from this. What to do after these steps.

Check out this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... at-part-i/

A number can be written in the form a^2 - b^2 if it is odd or has 4 as a factor (explained in the post above)

There are 4C2 = 6 ways of picking a pair of numbers here. Out of these, in only two cases (2, 5) and (2, 7) you will not be able to write the product as a^2 - b^2. Since product will not be odd and will not have 4 as a factor.
So required probability = 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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.

The products are 10, 14, 16, 35, 40, 56
The point to note is a^2 has to be bigger than the product and b^2 lesser than the product. So we use value of b whose square is less than the product
a^2=10 + b^2. we can easily see this is not a solution
a^2= 14 + b^2. . This can also be easily ruled out
a^2= 16 + b^2. for b =3 solution happens
a^2 = 35 + b^2. . for b=1 solution happens
a^2 = 40 +b ^2. for b=3 solution happens
a^2= 56 + b^2. for b =5 solution happens

probability = 4/6=2/3
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If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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.

As the number of terms is small, let's write down all possible outcomes

10, 14, 16, 35, 40, 56

We want the product to be of the form $$a^2 - b^2$$ = $$(a+b)(a-b)$$

(a+b) & (a-b) can be visualized as two points on the number lines which are b units away and lie on the right of a and left of a respectively. Hence, 'a' lies at the center of (a+b) and (a-b) and can be expressed by $$\frac{(a+b)+(a-b)}{2}$$. It's given that 'a' and 'b' are integers.

Attachment:

Screenshot 2023-04-01 152202.jpg [ 22.69 KiB | Viewed 799 times ]

Inference: Both (a+b) and (a-b) must have the same even-odd nature. This is required as (a+b) + (a-b) MUST be even for the sum to be divisible by 2.

With some pre-analysis done, let's factorize each of the outcomes -

10

5 * 2
10 * 1

Both factors don't hold the same even-odd nature. Hence, 10 is not a favorable number

Note: The midpoint 'a' is not an integer in either of the cases

5 * 2 ⇒ (5+2)/2 = 3.5
10 * 1 ⇒ (10 + 1) / 2 = 5.5

14

7 * 2
15 * 1

Both factors don't hold the same even-odd nature. Hence, 14 is not a favorable number

16

$$2^4$$

1 * 16
2 * 8
4 * 4

Both 2 and 8 are even, hence the midpoint of 2 and 8 will be an integer.

a = $$\frac{2 + 8 }{ 2}$$ = 5 and we can express the product as (5+3)(5-3)

16 is a favorable number

35

1 * 35
7 * 5

1 & 35 and 7 & 5 are odd pairs, hence the midpoint of each pair will be an integer.

a = $$\frac{7 + 5 }{ 2}$$ = 12 and we can express the product as (6+1)(6-1)

35 is a favorable number

40

1 * 40
2 * 20
4 * 10

2 * 20 & 4 * 10 are even pairs, hence the midpoint of each pair will be an integer.

a = $$\frac{2 + 20 }{ 2}$$ = 11 and we can express the product as (11+9)(11-9)

40 is a favorable number

56

1 * 56
7 * 8
14 * 4

14 & 4 is an even pair, hence the midpoint will be an integer.

a = $$\frac{14 + 4 }{ 2}$$ = 9 and we can express the product as (9+5)(9-5)

56 is a favorable number
Note: We need not check for all pairs of factors. If we are able to figure out that at least one pair of factors share the same even-odd nature, we can consider the number favorable.

Favorable Pairs: 4

Total Pairs: 6

Required Probability = 4/6 = 2/3

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

We need to express the product as a^2 - b^2 = (a + b)(a - b)
Thus, (a + b) and (a - b) will be factors of the product. For this to be possible, (a + b) and (a - b) should both be odd or both be even
The only cases where this is not possible is if one of them is odd and the other even.

Why? ... Say:
a + b = Odd
a - b = Even
Adding: 2a = Odd => a is NOT an integer - Hence it is not possible

From the numbers {2, 5, 7, 8}, the combinations where one will be even and the other odd will be when the number formed is 2 * Odd
For example: 2 * 5 and 2 * 7
In all the other cases, we can split the numbers as even * even or odd * odd.
For example, if we chose 8 * 5, we get 40, which is 10 * 4 => a + b = 10 and a - b = 4 => a = 7 and b = 3; and so on

Total number of ways of choosing 2 numbers of the 4 is 4C2 = 6
Of these, only 2 cases (2*5 and 2*7) are not possible

Thus, probability = 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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.

The consecutive differences between Perfect Squares are the odd numbers beginning from 3, so any odd product can be written in a difference between two Perfect Squares as 5×3, which is the only odd product. If the product is even, it must be divisible by 4. Because if you divide a multiple of 4 by 2, gives you an even number, this even number is between two odd consecutive numbers, which their sum is a difference between two Perfect Squares, but not consecutive.

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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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