Alan

"Ramanujan and Hardy played a game where they both picked a complex number. If the product of their numbers was 17 - 20i, and Hardy picked 7 - 2i, what number did Ramanujan pick?"

Let Ramanujan number be a + bi, then

(a + bi)*(7 - 2i) = 17 - 20i

7a - 2ai + 7bi + b = 17 - 20i

Equate real and imaginary parts:

7a + b = 17

-2a + 7b = -20

Solve these two equations for a and b.

Like so:

Let \(u = 2^x\), then we have \(u^2+u+1=111\), or \(u^2 + u - 110 = 0 \) or \((u - 10)(u + 11) = 0\)

Assuming ony a real number solution for x means we want u = 10 and \(x=\frac{\ln(10)}{\ln(2)}\) 

"The scientists captured and tagged 500 fish and then released them into the lake. Later, the scientists captured 500 more fish and counted that 4 of them were tagged. How many fish can the scientists estimate are in the lake?"

This should help:

If n = 3 then

mod(n^2,5)  =  4

mod(n^3,5)  =  2

mod(n , 5)  =   What do you think?

As follows:

But, more importantly, do you understand why these are the choices?

But x=1/2 and y=1/2 leads to x-y = 0.

Try x = 1/2 and y = 0.   We still have 2(x^2 + y^2) = x + y, but now x - y = 1/2.

As follows:

Presumably, the equation is 4x² + nx + 49 = 0.

This has a single solution when the discriminant is zero; that is when \(n^2-4*4*49=0\)

Can you take it from here?