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01 Jun 2018, 18:22
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. \(2\)



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20 Jul 2018, 23:01
Did everything correct except the last step to arrange in ascending order, in hurry! and thus ended up with wrong ans 0. Silly mistake.



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25 Jul 2018, 14:20
Can you anyone share example of a number that is not a perfect square and has only 2 factors?



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25 Jul 2018, 14:20
gsingh0711 wrote: Can you anyone share example of a number that is not a perfect square and has only 2 factors? Bunuel



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25 Jul 2018, 20:23
gsingh0711 wrote: gsingh0711 wrote: Can you anyone share example of a number that is not a perfect square and has only 2 factors? Bunuel1. All primes have two factors: 2, 3, 5, 7, ... 2. No perfect square has two factors.
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26 Jul 2018, 16:17
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? Bunuel1. 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?



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26 Jul 2018, 20:30
gsingh0711 wrote: Bunuel.. Thanks for the reply. Are we not suppose to consider negative factors we well? Factor = positive divisor.
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13 Sep 2018, 01:54
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 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 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 nonperfect 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 backcalculating 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 nonperfect square integers' Right?? Please confirm. TIA



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11 Jan 2019, 23:38
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



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11 Jan 2019, 23:49
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 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 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



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12 Jan 2019, 03:29
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 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 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.
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12 Jan 2019, 13:10
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 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 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



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13 Feb 2019, 09:25
I think this is a highquality question and I agree with explanation.
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