Puzzles

How to get started with this puzzle is easiest to see if we plot the two curves:

We see that the two curves meet at the point (2,3). The only way that the straight line \(f\) can be between the two curves is if it is a tangent to both curves.

By differentiation of another way of finding tangents, you can show that \(a=2\) and \(b=1\). Therefore \(f(200)\) is 399.

A line \(y=\text{constant}\) that is tangent to a parabola must be a tangent to the minimum (or maximum) point.

\(y=x^2-ax\) can be rewritten as \(y=(x-a/2)^2-a^2/4\). The minimum of this will be at \(y=-a^2/4\). If \(-a^2/4=-160\,000\), then \(a\) is 800.

If we plot the two quadratics, we see that they meet at \(x=2\) where they both have gradient 4.

The line \(f(x)\) must therefore pass through the point \((2,4)\) and have gradient 4, so its equation is \(f(x)=4x-4\).

This means that \(f(65)\) is 256.

If the equation has two integer solutions, then it can be written as \((x+\alpha)(x+\beta)\), where \(\alpha\) and \(\beta\) are integers. Expanding this and setting it equal to \(x^2+bx+414720=0\), we find that \(b=\alpha+\beta\) and \(414720=\alpha\beta\).

414720 has 55 different pairs of factors. Each of these pairs leads to two values of \(b\) (a positive and a negative value). Therefore there are 110 values of \(b\) that give the equation two integer solutions.

If \(x^2-7x+11=1\) or \(x^2-11x+30=0\), then this is one. The solutions to these are \(x=2,5,\) and \(6\).

It could also be one if \(x^2-7x+11=-1\) and \(x^2-11x+30\) is even. This happens when \(x=3\) or \(4\).

Therefore all the solutions to this are \(x=2,3,4,5\) or \(6\).