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# If two integers are chosen at random out of first 5 positive integers,

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Director
Joined: 24 Oct 2016
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If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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11 May 2019, 06:06
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If two integers are chosen at random out of first 5 positive integers, 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. 2/5
B. 3/5
C. 7/10
D. 4/5
E. 9/10
Director
Joined: 24 Oct 2016
Posts: 585
GMAT 1: 670 Q46 V36
GMAT 2: 690 Q47 V38
Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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11 May 2019, 06:08
3
8
dabaobao wrote:
If two integers are chosen at random out of first 5 positive integers, 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. 2/5
B. 3/5
C. 7/10
D. 4/5
E. 9/10

TL;DR

#s: 1, 2, 3, 4, 5

Total Outcome: 5C2 = 5!/3!2! = 10
Favourable Cases:

n1.n2 = a^2 - b^2 = (a+b)(a-b)

If n1=2 (3.5-1.5) and n2=5 (3.5+1.5), then a=3.5 b=1.5 => Not possible since a & b are + int

For a & b to be int, avg needs to be int, which is possible only when both chosen numbers are even or odd
#even: 2 => Fav case = 1
#odd: 3 => Fav case = 3C2 = 6
Total Fav case: 7 => ANSWER = 7/10

Official Solution

Solution: This might look like a probability question but isn’t. Questions like these are the reason we ask you to go through basics of every topic including probability. If you do not know probability at all, you may skip this question even though it needs very basic knowledge of probability.

Probability will tell you that

Required probability = Favorable cases/Total cases

Total cases are very easy to find: 5C2 = 10 or 5*4/2 = 10 whatever you prefer. This is the number of ways in which you select any 2 distinct numbers out of the given 5 distinct numbers.

Number of favorable cases is the challenge here. That is why it is a number properties question and not so much a probability question. Let’s focus on the main part of the question:

First five positive integers: 1, 2, 3, 4, 5

We need to select two integers such that their product is of the form a^2 – b^2. What does a^2 – b^2 remind you of? It reminds me of (a + b)(a – b). So the product needs to be of the form (a + b)(a – b). So is it necessary that of the two numbers we pick, one must be of the form (a + b) and the other must be (a – b)? No. Note that we should be able to write the product in this form. It is not necessary that the numbers must be of this form only.

But first let’s focus on numbers which are already of the form (a + b) and (a – b).

Say you pick two numbers, 2 and 5. Are they of the form (a + b) and (a – b) such that a and b are integers? No.

5 = 3.5 + 1.5

2 = 3.5 – 1.5

So a = 3.5, b = 1.5.

a and b are not integers.

What about numbers such as 3 and 5? Are they of the form (a + b) and (a – b) such that a and b are integers? Yes.

5 = 4 + 1

3 = 4 – 1

Note that whenever the average of the numbers will be an integer, we will be able to write them as a+b and a – b because one number will be some number more than the average and the other will be the same number less than average. So a will be the average and the amount more or less will be b.

When will the average of two numbers (Number1 + Number2)/2 be an integer? When the sum of the two numbers is even! When is the sum of two numbers even? It is when both the numbers are even or when both are odd. So then does the question boil down to “favorable cases are when we select both numbers even or both numbers odd?” Yes and No. When we select both even numbers or both odd numbers, the product can be written as a^2 – b^2. But are those the only cases when the product can be written as a^2 – b^2?

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. We need to be able to write the product (whatever we obtain) as product of two even or two odd numbers.

To explain this, let’s say we pick two numbers 4 and 5

4*5 = 20

Can we write 20 as product of two even numbers? Yes 2*10.

So even though, 4 is even and 5 is odd, their product can be written as product of two even numbers. So in which all cases will this happen?

- 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.

Let’s go back to our question:

We have 5 numbers: 1, 2, 3, 4, 5

Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor.

3 Odd numbers: 1, 3, 5

2 Even numbers: 2, 4

Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers)

Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4

Total number of favorable cases = 3 + 4 = 7

Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor.

Required Probability = 7/10

Takeaway:

When can we write a number as difference of squares?

- When the number is odd

or

- When the number has 4 as a factor
##### General Discussion
Intern
Joined: 24 Mar 2019
Posts: 45
If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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11 May 2019, 07:24
1
Please explain how are we selecting a combination of (1,4)-->the product of this i.e 4 cannot be represented in the form of (a+b)(a-b) ,where a and b are both +ve integers.

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Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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24 May 2019, 04:31
2
Total # of ways: 5C2 = 10
Eligible choices:
(1) 1*3=4-1
(2) 1*5=9-4
(3) 2*4=9-1
(4) 3*4=16-4
(5) 3*5=16-1
(6) 4*5=36-16

What's the 7th eligible choice?
VP
Joined: 19 Oct 2018
Posts: 1308
Location: India
Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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24 May 2019, 06:18
1
Bunuel (1*2), (1*4) (2*3) and (2*5) can't be written in the form of (a^2)-(b^2), if and b are positive integers. Answer should be B.
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Posts: 2012
Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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24 May 2019, 06:55
3
1
Yes, there are several problems with this question - while 4 can be written in the form a^2 - b^2 using integers, you can only do that if you use zero (e.g. 4 = 2^2 - 0^2), which the question disallows. So the official solution is counting one case it should not be counting, and the official answer is incorrect. The question also needs to specify if we're making selections with or without replacement. And of course any problem that requires a 20-paragraph solution is not a realistic GMAT question.
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Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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24 Jun 2019, 12:55
dabaobao wrote:
dabaobao wrote:
If two integers are chosen at random out of first 5 positive integers, 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. 2/5
B. 3/5
C. 7/10
D. 4/5
E. 9/10

TL;DR

#s: 1, 2, 3, 4, 5

Total Outcome: 5C2 = 5!/3!2! = 10
Favourable Cases:

n1.n2 = a^2 - b^2 = (a+b)(a-b)

If n1=2 (3.5-1.5) and n2=5 (3.5+1.5), then a=3.5 b=1.5 => Not possible since a & b are + int

For a & b to be int, avg needs to be int, which is possible only when both chosen numbers are even or odd
#even: 2 => Fav case = 1
#odd: 3 => Fav case = 3C2 = 6
Total Fav case: 7 => ANSWER = 7/10

Official Solution

Solution: This might look like a probability question but isn’t. Questions like these are the reason we ask you to go through basics of every topic including probability. If you do not know probability at all, you may skip this question even though it needs very basic knowledge of probability.

Probability will tell you that

Required probability = Favorable cases/Total cases

Total cases are very easy to find: 5C2 = 10 or 5*4/2 = 10 whatever you prefer. This is the number of ways in which you select any 2 distinct numbers out of the given 5 distinct numbers.

Number of favorable cases is the challenge here. That is why it is a number properties question and not so much a probability question. Let’s focus on the main part of the question:

First five positive integers: 1, 2, 3, 4, 5

We need to select two integers such that their product is of the form a^2 – b^2. What does a^2 – b^2 remind you of? It reminds me of (a + b)(a – b). So the product needs to be of the form (a + b)(a – b). So is it necessary that of the two numbers we pick, one must be of the form (a + b) and the other must be (a – b)? No. Note that we should be able to write the product in this form. It is not necessary that the numbers must be of this form only.

But first let’s focus on numbers which are already of the form (a + b) and (a – b).

Say you pick two numbers, 2 and 5. Are they of the form (a + b) and (a – b) such that a and b are integers? No.

5 = 3.5 + 1.5

2 = 3.5 – 1.5

So a = 3.5, b = 1.5.

a and b are not integers.

What about numbers such as 3 and 5? Are they of the form (a + b) and (a – b) such that a and b are integers? Yes.

5 = 4 + 1

3 = 4 – 1

Note that whenever the average of the numbers will be an integer, we will be able to write them as a+b and a – b because one number will be some number more than the average and the other will be the same number less than average. So a will be the average and the amount more or less will be b.

When will the average of two numbers (Number1 + Number2)/2 be an integer? When the sum of the two numbers is even! When is the sum of two numbers even? It is when both the numbers are even or when both are odd. So then does the question boil down to “favorable cases are when we select both numbers even or both numbers odd?” Yes and No. When we select both even numbers or both odd numbers, the product can be written as a^2 – b^2. But are those the only cases when the product can be written as a^2 – b^2?

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. We need to be able to write the product (whatever we obtain) as product of two even or two odd numbers.

To explain this, let’s say we pick two numbers 4 and 5

4*5 = 20

Can we write 20 as product of two even numbers? Yes 2*10.

So even though, 4 is even and 5 is odd, their product can be written as product of two even numbers. So in which all cases will this happen?

- 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.

Let’s go back to our question:

We have 5 numbers: 1, 2, 3, 4, 5

Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor.

3 Odd numbers: 1, 3, 5

2 Even numbers: 2, 4

Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers)

Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4

Total number of favorable cases = 3 + 4 = 7

Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor.

Required Probability = 7/10

Takeaway:

When can we write a number as difference of squares?

- When the number is odd

or

- When the number has 4 as a factor

how did u write
#odd: 3 => Fav case = 3C2 = 6
pls explain
VP
Joined: 19 Oct 2018
Posts: 1308
Location: India
Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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24 Jun 2019, 13:09
His solution is wrong, and even official answer is incorrect. But concept used to solve this question is a good one.

Neha2050 wrote:
dabaobao wrote:
dabaobao wrote:
If two integers are chosen at random out of first 5 positive integers, 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. 2/5
B. 3/5
C. 7/10
D. 4/5
E. 9/10

TL;DR

#s: 1, 2, 3, 4, 5

Total Outcome: 5C2 = 5!/3!2! = 10
Favourable Cases:

n1.n2 = a^2 - b^2 = (a+b)(a-b)

If n1=2 (3.5-1.5) and n2=5 (3.5+1.5), then a=3.5 b=1.5 => Not possible since a & b are + int

For a & b to be int, avg needs to be int, which is possible only when both chosen numbers are even or odd
#even: 2 => Fav case = 1
#odd: 3 => Fav case = 3C2 = 6
Total Fav case: 7 => ANSWER = 7/10

Official Solution

Solution: This might look like a probability question but isn’t. Questions like these are the reason we ask you to go through basics of every topic including probability. If you do not know probability at all, you may skip this question even though it needs very basic knowledge of probability.

Probability will tell you that

Required probability = Favorable cases/Total cases

Total cases are very easy to find: 5C2 = 10 or 5*4/2 = 10 whatever you prefer. This is the number of ways in which you select any 2 distinct numbers out of the given 5 distinct numbers.

Number of favorable cases is the challenge here. That is why it is a number properties question and not so much a probability question. Let’s focus on the main part of the question:

First five positive integers: 1, 2, 3, 4, 5

We need to select two integers such that their product is of the form a^2 – b^2. What does a^2 – b^2 remind you of? It reminds me of (a + b)(a – b). So the product needs to be of the form (a + b)(a – b). So is it necessary that of the two numbers we pick, one must be of the form (a + b) and the other must be (a – b)? No. Note that we should be able to write the product in this form. It is not necessary that the numbers must be of this form only.

But first let’s focus on numbers which are already of the form (a + b) and (a – b).

Say you pick two numbers, 2 and 5. Are they of the form (a + b) and (a – b) such that a and b are integers? No.

5 = 3.5 + 1.5

2 = 3.5 – 1.5

So a = 3.5, b = 1.5.

a and b are not integers.

What about numbers such as 3 and 5? Are they of the form (a + b) and (a – b) such that a and b are integers? Yes.

5 = 4 + 1

3 = 4 – 1

Note that whenever the average of the numbers will be an integer, we will be able to write them as a+b and a – b because one number will be some number more than the average and the other will be the same number less than average. So a will be the average and the amount more or less will be b.

When will the average of two numbers (Number1 + Number2)/2 be an integer? When the sum of the two numbers is even! When is the sum of two numbers even? It is when both the numbers are even or when both are odd. So then does the question boil down to “favorable cases are when we select both numbers even or both numbers odd?” Yes and No. When we select both even numbers or both odd numbers, the product can be written as a^2 – b^2. But are those the only cases when the product can be written as a^2 – b^2?

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. We need to be able to write the product (whatever we obtain) as product of two even or two odd numbers.

To explain this, let’s say we pick two numbers 4 and 5

4*5 = 20

Can we write 20 as product of two even numbers? Yes 2*10.

So even though, 4 is even and 5 is odd, their product can be written as product of two even numbers. So in which all cases will this happen?

- 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.

Let’s go back to our question:

We have 5 numbers: 1, 2, 3, 4, 5

Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor.

3 Odd numbers: 1, 3, 5

2 Even numbers: 2, 4

Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers)

Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4

Total number of favorable cases = 3 + 4 = 7

Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor.

Required Probability = 7/10

Takeaway:

When can we write a number as difference of squares?

- When the number is odd

or

- When the number has 4 as a factor

how did u write
#odd: 3 => Fav case = 3C2 = 6
pls explain
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Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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05 Jul 2019, 21:17
The question can be worded slightly better: The probability of picking two positive integers a and b from first 5 positive integers so that their product can be represented in the general form p^2-q^2 where p and q share the same set as a and b

if you consider that then you limit your set to first 5 positive integers:

5C2 ways of picking two numbers

3*1= 4-1
5*1=9-4
4*2= 9-1
4*3= 16-4
5*3= 16-1
4*4= 25-9
3*3= 25-16

Please see that are numbers are picked from the set {1,2,3,4,5}

7 cases possible hence 7/10
VP
Joined: 19 Oct 2018
Posts: 1308
Location: India
Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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06 Jul 2019, 02:22
You can't select 2 4's from first 5 positive integers. That's wrong imo.
vinayakvaish wrote:
The question can be worded slightly better: The probability of picking two positive integers a and b from first 5 positive integers so that their product can be represented in the general form p^2-q^2 where p and q share the same set as a and b

if you consider that then you limit your set to first 5 positive integers:

5C2 ways of picking two numbers

3*1= 4-1
5*1=9-4
4*2= 9-1
4*3= 16-4
5*3= 16-1
4*4= 25-9
3*3= 25-16

Please see that are numbers are picked from the set {1,2,3,4,5}

7 cases possible hence 7/10

Posted from my mobile device
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Posts: 114
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GMAT 2: 720 Q49 V39
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Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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06 Jul 2019, 05:33
nick1816 wrote:
You can't select 2 4's from first 5 positive integers. That's wrong imo.
vinayakvaish wrote:
The question can be worded slightly better: The probability of picking two positive integers a and b from first 5 positive integers so that their product can be represented in the general form p^2-q^2 where p and q share the same set as a and b

if you consider that then you limit your set to first 5 positive integers:

5C2 ways of picking two numbers

3*1= 4-1
5*1=9-4
4*2= 9-1
4*3= 16-4
5*3= 16-1
4*4= 25-9
3*3= 25-16

Please see that are numbers are picked from the set {1,2,3,4,5}

7 cases possible hence 7/10

Posted from my mobile device
c

Unless otherwise stated, you can not assume this in GMAT. I have seen so many questions where people miss a case because they fail to consider this. For questions requiring this, the stem reads "x different integers"

I hope it helps
VP
Joined: 19 Oct 2018
Posts: 1308
Location: India
Re: If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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06 Jul 2019, 05:56

Why did you not consider those cases in total number of ways to select 2 integers from 5 first five positive integers.
According to your logic- total number of ways to pick 2 numbers out of 5 numbers = 5C2+5=15

(1,1), (1,2), (1,3), (1,4), (1,5), (2,2), (2,3), (2,4), (2,5), (3,3), (3,4), (3,5), (4,4), (4,5) and (5,5)

vinayakvaish wrote:
nick1816 wrote:
You can't select 2 4's from first 5 positive integers. That's wrong imo.
vinayakvaish wrote:
The question can be worded slightly better: The probability of picking two positive integers a and b from first 5 positive integers so that their product can be represented in the general form p^2-q^2 where p and q share the same set as a and b

if you consider that then you limit your set to first 5 positive integers:

5C2 ways of picking two numbers

3*1= 4-1
5*1=9-4
4*2= 9-1
4*3= 16-4
5*3= 16-1
4*4= 25-9
3*3= 25-16

Please see that are numbers are picked from the set {1,2,3,4,5}

7 cases possible hence 7/10

Posted from my mobile device
c

Unless otherwise stated, you can not assume this in GMAT. I have seen so many questions where people miss a case because they fail to consider this. For questions requiring this, the stem reads "x different integers"

I hope it helps
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If two integers are chosen at random out of first 5 positive integers,  [#permalink]

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18 Jul 2019, 22:46
I think for the official answer given, the question would have been correct, had it stated that a and b as just integers, instead of positive integers.

If we were to choose numbers 4,1 from first5 positive integers, then their product 4 can be written as $$a^2 - b^2$$ only if a = 2 and b = 0, (but actual question a and b are positive integers).

if a, b are to be +ve integers, then (4,1) cannot be combination, so effectively, only 6 combinations satisfy the given condition, so the answer must be (6/10) = (3/5) which is B, IMO
If two integers are chosen at random out of first 5 positive integers,   [#permalink] 18 Jul 2019, 22:46
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