trumpstinks

\(QED = (11-5i)(3-2i)(2i)\)

We can split this up into two parts. Let's first compute \((11-5i)(3-2i)\), and then multiply it again by \(2i\).

\((11-5i)(3-2i) = 11(3-2i)-5i(3-2i) = 33-22i-15i+10i^2 = 33-37i-10 = 23-37i.\)

\(2i\cdot(23-37i) = 46i-74i^2 = \boxed{74+46i}\)

jimkey17, you had the right idea, but you made a mistake in your work. It should be choose 5, not 4. The real answer is 11, as 11 choose 4 > 365, but 10 choose 4 isn't. (Note: I solved this problem on AoPS, and 11 is correct, so you need have no doubts in the solution)  

I already tried that response a long time ago. It was wrong. I'm confused.

Since \(p\), \(q\), and \(r\) must all be positive, \(p = 1\) because if it was any higher, \(q\) and \(r\) would need to be negative to make equation true. So we have \(1+\frac{1}{q+\frac{1}{r}}=\frac{25}{19}\). So \(\frac{1}{q+\frac{1}{r}}=\frac{6}{19}= \frac{6}{6q+\frac{6}{r}}\), so \(19=6q+\frac{6}{r}\). The only possibility of \(q\) is \(\boxed{3}\), with \(r\) being \(6\).

thanks!! I thought it was only integers ;(