M30-12

Official Solution:


The product of two negative integers, \(a\) and \(b\), is a prime number \(p\). If \(p\) is the number of factors of a positive integer \(n\), 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 question covers several concepts in number theory and is quite challenging.

Let's begin with the positive integer \(n\), which we know is not a perfect square. We can recall that positive perfect squares have an odd number of factors while all other positive integers have an even number of factors. Hence, the number of factors of \(n\), which is \(p\), must be even. We are also told that \(p\) is a prime number, and since the only even prime number is \(2\), it follows that \(p=2\). Moreover, since only prime numbers have exactly \(2\) factors (1 and itself), \(n\) must also be a prime number.

Given that \(ab=p=2\), we have either \(a=-1\) and \(b=-2\), or vice versa. Therefore, the set of four integers is {-2, -1, 2, some prime}. The median of this set is \(\frac{-1+2}{2}=\frac{1}{2}\).


Answer: B