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Intern  S
Joined: 29 Jan 2017
Posts: 44

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I found the question to be fairly easy by picking numbers, but I'm starting to think if I just got lucky in terms of the number I picked. I struggle the most with number theory/properties

First I chose a number that fit the criteria for n; n=5
n=2
a and b = -1 and -2 in either order

Bunuel wrote:
The product of two negative integers, $$a$$ and $$b$$, is a prime number $$p$$. If $$p$$ is the number of factors of $$n$$, where $$n$$ is NOT a perfect square, what is the value of the median of the four integers $$a$$, $$b$$, $$p$$, and $$n$$?

A. $$0$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{3}{2}$$
E. $$2$$
Intern  B
Joined: 20 Aug 2016
Posts: 49
GMAT 1: 570 Q46 V23 GMAT 2: 610 Q49 V25 GMAT 3: 620 Q45 V31 WE: Information Technology (Other)

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Did everything correct except the last step to arrange in ascending order, in hurry! and thus ended up with wrong ans 0.
Silly mistake.
Manager  S
Joined: 02 Mar 2018
Posts: 67
Location: India
GMAT 1: 640 Q51 V26 GPA: 3.1

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Can you anyone share example of a number that is not a perfect square and has only 2 factors?
Manager  S
Joined: 02 Mar 2018
Posts: 67
Location: India
GMAT 1: 640 Q51 V26 GPA: 3.1

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gsingh0711 wrote:
Can you anyone share example of a number that is not a perfect square and has only 2 factors?

Bunuel
Math Expert V
Joined: 02 Sep 2009
Posts: 53865

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gsingh0711 wrote:
gsingh0711 wrote:
Can you anyone share example of a number that is not a perfect square and has only 2 factors?

Bunuel

1. All primes have two factors: 2, 3, 5, 7, ...
2. No perfect square has two factors.
_________________
Manager  S
Joined: 02 Mar 2018
Posts: 67
Location: India
GMAT 1: 640 Q51 V26 GPA: 3.1

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Bunuel wrote:
gsingh0711 wrote:
gsingh0711 wrote:
Can you anyone share example of a number that is not a perfect square and has only 2 factors?

Bunuel

1. All primes have two factors: 2, 3, 5, 7, ...
2. No perfect square has two factors.

Bunuel.. Thanks for the reply. Are we not suppose to consider negative factors we well?
Math Expert V
Joined: 02 Sep 2009
Posts: 53865

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gsingh0711 wrote:

Bunuel.. Thanks for the reply. Are we not suppose to consider negative factors we well?

Factor = positive divisor.
_________________
Intern  B
Joined: 30 Apr 2018
Posts: 19

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Bunuel wrote:
Official Solution:

The product of two negative integers, $$a$$ and $$b$$, is a prime number $$p$$. If $$p$$ is the number of factors of $$n$$, where $$n$$ is NOT a perfect square, what is the value of the median of the four integers $$a$$, $$b$$, $$p$$, and $$n$$?

A. $$0$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{3}{2}$$
E. $$2$$

This is a hard questions which tests several number theory concepts.

Start from $$n$$: we are told that $$n$$ is NOT a perfect square. The number of factors of perfect square is odd and all other positive integers have even number of factors. Hence, since $$p$$ is the number of factors of $$n$$, then $$p$$ must be even. We also know that $$p$$ is a prime number and since the only even prime is 2, then $$p=2$$. Notice here that from this it follows that $$n$$ must also be a prime, because only primes have 2 factors: 1 and itself.

Next, $$ab=p=2$$ implies that $$a=-1$$ and $$b=-2$$ or vise-versa.

So, the set is {-2, -1, 2, some prime}, which means that the median is $$\frac{-1 + 2}{2} = \frac{1}{2}$$.

Hi Bunuel,

This is a great solution.

I was wondering - if at all the solution is possible without knowing the concept around no. of factors for perfect square integers and non-perfect square integers - My try at it says, it is not possible and i wanted to confirm it with an expert.

The way i tried solving it (assuming that one wouldnt know the '# of factors' concept)

1. since a prime no. can be product of two nos. only when they are its factors hence between a and b, one has to be 1 and one has to be the prime no. itself
2. since both are negative and prime nos. are always positive, hence a,b will have to be {-1, -p (i.e. the prime no. p)}
Also as the smallest prime n is = 2, hence between a and b, one will always be more negative than -1, and hence smaller
3. since median for a list of 4 nos. = (2nd no + 3rd no)/2, arranged in ascending order;
for the list {-p, -1, p, some number n} => the median = (-1+p)/2 => p = (2*median)+1
4. Now, that i start back-calculating for p, by using the answer options, the option where p = a prime no. should be the correct answer
5. As both options B and D result in a prime no - i.e. 2 and 5; to be able to choose from one of them - the only way forward is to know the 'the concept around no. of factors for perfect square integers and non-perfect square integers'

Right??

Intern  B
Joined: 22 Sep 2016
Posts: 6
Location: Cameroon
Schools: INSEAD Jan '19
GPA: 3.21

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Bunuel wrote:
The product of two negative integers, $$a$$ and $$b$$, is a prime number $$p$$. If $$p$$ is the number of factors of $$n$$, where $$n$$ is NOT a perfect square, what is the value of the median of the four integers $$a$$, $$b$$, $$p$$, and $$n$$?

A. $$0$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{3}{2}$$
E. $$2$$

Hello Brunnel,
you wrote that "The number of factors of perfect square is odd and all other positive integers have even number of factors." 6 is not a perfect square but has 3 factors. Do I miss something?
Could you please shed light on this point?

Regards
Yanick
Intern  B
Joined: 22 Sep 2016
Posts: 6
Location: Cameroon
Schools: INSEAD Jan '19
GPA: 3.21

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Bunuel wrote:
Official Solution:

The product of two negative integers, $$a$$ and $$b$$, is a prime number $$p$$. If $$p$$ is the number of factors of $$n$$, where $$n$$ is NOT a perfect square, what is the value of the median of the four integers $$a$$, $$b$$, $$p$$, and $$n$$?

A. $$0$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{3}{2}$$
E. $$2$$

This is a hard questions which tests several number theory concepts.

Start from $$n$$: we are told that $$n$$ is NOT a perfect square. The number of factors of perfect square is odd and all other positive integers have even number of factors. Hence, since $$p$$ is the number of factors of $$n$$, then $$p$$ must be even. We also know that $$p$$ is a prime number and since the only even prime is 2, then $$p=2$$. Notice here that from this it follows that $$n$$ must also be a prime, because only primes have 2 factors: 1 and itself.

Next, $$ab=p=2$$ implies that $$a=-1$$ and $$b=-2$$ or vise-versa.

So, the set is {-2, -1, 2, some prime}, which means that the median is $$\frac{-1 + 2}{2} = \frac{1}{2}$$.

Hello Brunnel,
you wrote that "The number of factors of perfect square is odd and all other positive integers have even number of factors."
6 is not a perfect square but has 3 factors.
Do I miss something?
Could you please shed light on this point?

Regards
Yanick
Math Expert V
Joined: 02 Sep 2009
Posts: 53865

### Show Tags

Yanick wrote:
Bunuel wrote:
Official Solution:

The product of two negative integers, $$a$$ and $$b$$, is a prime number $$p$$. If $$p$$ is the number of factors of $$n$$, where $$n$$ is NOT a perfect square, what is the value of the median of the four integers $$a$$, $$b$$, $$p$$, and $$n$$?

A. $$0$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{3}{2}$$
E. $$2$$

This is a hard questions which tests several number theory concepts.

Start from $$n$$: we are told that $$n$$ is NOT a perfect square. The number of factors of perfect square is odd and all other positive integers have even number of factors. Hence, since $$p$$ is the number of factors of $$n$$, then $$p$$ must be even. We also know that $$p$$ is a prime number and since the only even prime is 2, then $$p=2$$. Notice here that from this it follows that $$n$$ must also be a prime, because only primes have 2 factors: 1 and itself.

Next, $$ab=p=2$$ implies that $$a=-1$$ and $$b=-2$$ or vise-versa.

So, the set is {-2, -1, 2, some prime}, which means that the median is $$\frac{-1 + 2}{2} = \frac{1}{2}$$.

Hello Brunnel,
you wrote that "The number of factors of perfect square is odd and all other positive integers have even number of factors."
6 is not a perfect square but has 3 factors.
Do I miss something?
Could you please shed light on this point?

Regards
Yanick

6 = 2*3 has (1 + 1)(1 + 1) = 4 factors: 1, 2, 3, and 6.
_________________
Intern  B
Joined: 22 Sep 2016
Posts: 6
Location: Cameroon
Schools: INSEAD Jan '19
GPA: 3.21

### Show Tags

Bunuel wrote:
Yanick wrote:
Bunuel wrote:
Official Solution:

The product of two negative integers, $$a$$ and $$b$$, is a prime number $$p$$. If $$p$$ is the number of factors of $$n$$, where $$n$$ is NOT a perfect square, what is the value of the median of the four integers $$a$$, $$b$$, $$p$$, and $$n$$?

A. $$0$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{3}{2}$$
E. $$2$$

This is a hard questions which tests several number theory concepts.

Start from $$n$$: we are told that $$n$$ is NOT a perfect square. The number of factors of perfect square is odd and all other positive integers have even number of factors. Hence, since $$p$$ is the number of factors of $$n$$, then $$p$$ must be even. We also know that $$p$$ is a prime number and since the only even prime is 2, then $$p=2$$. Notice here that from this it follows that $$n$$ must also be a prime, because only primes have 2 factors: 1 and itself.

Next, $$ab=p=2$$ implies that $$a=-1$$ and $$b=-2$$ or vise-versa.

So, the set is {-2, -1, 2, some prime}, which means that the median is $$\frac{-1 + 2}{2} = \frac{1}{2}$$.

Hello Brunnel,
you wrote that "The number of factors of perfect square is odd and all other positive integers have even number of factors."
6 is not a perfect square but has 3 factors.
Do I miss something?
Could you please shed light on this point?

Regards
Yanick

6 = 2*3 has (1 + 1)(1 + 1) = 4 factors: 1, 2, 3, and 6.

Bunuel.. Thanks for the reply. I got my mistake

Regards
Yanick
Manager  G
Joined: 22 Jun 2017
Posts: 178
Location: Argentina
Schools: HBS, Stanford, Wharton

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I think this is a high-quality question and I agree with explanation.
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The HARDER you work, the LUCKIER you get. Re M30-12   [#permalink] 13 Feb 2019, 09:25

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