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M30-12

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M30-12  [#permalink]

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New post 16 Sep 2014, 00:45
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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. \(2\)

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Re M30-12  [#permalink]

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New post 16 Sep 2014, 00:45
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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 \(\frac{-1 + 2}{2} = \frac{1}{2}\).


Answer: B
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Re M30-12  [#permalink]

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New post 10 Jan 2015, 23:55
I think this question is good and helpful.
Nice question covering 2-3 concepts
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Re: M30-12  [#permalink]

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New post 08 Apr 2015, 06:34
if the set is {-1,-2,2,some prime} then the median would be 0. Am I missing out something?
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Re: M30-12  [#permalink]

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New post 08 Apr 2015, 06:39
sachin2040 wrote:
if the set is {-1,-2,2,some prime} then the median would be 0. Am I missing out something?


The median of a set with even number of elements is the average of two middle terms when arranged in ascending /descending order. The set in ascending order is {-2, -1, 2, some prime} --> median = (-1 + 2)/2 = 1/2.
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Re: M30-12  [#permalink]

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New post 08 Apr 2015, 06:44
Thanks Bunuel, that was silly on my part.....
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Re: M30-12  [#permalink]

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New post 28 Apr 2016, 06:37
great concept question!
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Re: M30-12  [#permalink]

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New post 28 May 2016, 22:04
Notice here that from this it follows that n must also be a prime, because only primes have 2 factors: 1 and itself.

Can n be negative here? Say n = -1, its factors would be 1 and -1.
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Re: M30-12  [#permalink]

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Re: M30-12  [#permalink]

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New post 26 Jun 2016, 05:51
A very good question ! +1 Kudos :)
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Re M30-12  [#permalink]

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New post 01 Aug 2016, 13:09
I think this is a high-quality question and I agree with explanation.
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Re: M30-12  [#permalink]

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New post 04 Aug 2016, 16:25
Bunuel wrote:
sushruthav wrote:
Notice here that from this it follows that n must also be a prime, because only primes have 2 factors: 1 and itself.

Can n be negative here? Say n = -1, its factors would be 1 and -1.


No, a factor is a positive divisor.



Then, Suppose, 3 is the number of factors of N.
Can we say, N is a perfect Square?
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Re: M30-12  [#permalink]

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New post 05 Aug 2016, 03:30
nahid78 wrote:
Bunuel wrote:
sushruthav wrote:
Notice here that from this it follows that n must also be a prime, because only primes have 2 factors: 1 and itself.

Can n be negative here? Say n = -1, its factors would be 1 and -1.


No, a factor is a positive divisor.



Then, Suppose, 3 is the number of factors of N.
Can we say, N is a perfect Square?


Yes.

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

Hope it helps.
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Re: M30-12  [#permalink]

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New post 06 Oct 2016, 01:29
Bunuel wrote:
So, the set is {-2, -1, 2, some prime}, which means that the median is \(\frac{-1 + 2}{2} = \frac{1}{2}\). Answer: B


Didn't Get It!
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New post 06 Oct 2016, 03:53
RichaChampion wrote:
Bunuel wrote:
So, the set is {-2, -1, 2, some prime}, which means that the median is \(\frac{-1 + 2}{2} = \frac{1}{2}\). Answer: B


Didn't Get It!


The median is the average of two middle terms, when arranged in ascending/descending order. So, the median of {-2, -1, 2, some prime} is \(\frac{-1 + 2}{2} = \frac{1}{2}\) (regardless of the unknown prime there).
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Re: M30-12  [#permalink]

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New post 17 Nov 2016, 14:41
Bunuel wrote:
sushruthav wrote:
Notice here that from this it follows that n must also be a prime, because only primes have 2 factors: 1 and itself.

Can n be negative here? Say n = -1, its factors would be 1 and -1.


No, a factor is a positive divisor.


Could you please explain a bit more regarding factors of negative number? Can n be -2 in this case having two factors 1 and 2? From my understanding if number of factors of a number are given the number could be either positive or negative.
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Re: M30-12  [#permalink]

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New post 19 Nov 2016, 00:12
Great Question. Very helpful. +1 Kudos.
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New post 09 Jan 2017, 19:25
I think this is a high-quality question and I agree with explanation.
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Re: M30-12  [#permalink]

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New post 21 Apr 2018, 03:08
Since a and b can be -1 and -2 or vice versa, are 1/2 and 0 not a possible answer? How did we assume that a=-2 and b=-1?
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New post 21 Apr 2018, 04:16
prkohli wrote:
Since a and b can be -1 and -2 or vice versa, are 1/2 and 0 not a possible answer? How did we assume that a=-2 and b=-1?


Your question is not clear. How are you getting 0 or 1/2 for the median? What set do you have giving those values of median ?

We got that \(a=-1\) and \(b=-2\) or vise-versa.

So, the set is {-2, -1, p = 2, n = some prime}, which means that the median is (-1 + 2)/2 = 1/2.
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Re: M30-12 &nbs [#permalink] 21 Apr 2018, 04:16

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