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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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16 Sep 2014, 00:45
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 viseversa. 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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10 Jan 2015, 23:55
I think this question is good and helpful. Nice question covering 23 concepts



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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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08 Apr 2015, 06:44
Thanks Bunuel, that was silly on my part.....



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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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26 Jun 2016, 05:51
A very good question ! +1 Kudos
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01 Aug 2016, 13:09
I think this is a highquality question and I agree with explanation.



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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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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 Oddfactors, and EVEN number of Evenfactors. The reverse is also true: if a number has an ODD number of Oddfactors, and EVEN number of Evenfactors 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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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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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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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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19 Nov 2016, 00:12
Great Question. Very helpful. +1 Kudos.



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09 Jan 2017, 19:25
I think this is a highquality question and I agree with explanation.



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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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