MaxWong

Notice that \(n^5 + n^4 + 1 = (n^2 + n + 1)(n^3 - n + 1)\).

That means, for all \(n^2 + n + 1 \neq 1\) and \(n^3 - n + 1 \neq 1\), \(n^5 + n^4 + 1\) is composite.

This means when \(n^5 + n^4 + 1\) is prime, \(n^2 + n = 0\) or \(n^3 = n\)

Solving, we get n = -1 (rejected) or n = 0 (rejected) or n = 1.

When n = 1, \(n^5 + n^4 + 1 = 3\), which is a prime.

The only possibility is n = 1.

\(y' = \dfrac{y(t + 1) + (t + 1)^2}{t^2}\\ \dfrac{dy}{dt} = y\left(\dfrac1t + \dfrac1{t^2}\right) + \left(1 + \dfrac1t\right)^2\\ \dfrac{dy}{dt} - \left(\dfrac1t + \dfrac1{t^2}\right)y = \left(1 + \dfrac1t\right)^2\\ \)

We then find the integrating factor:

\(\mathcal{I}(t) = e^{-\int \left(\frac1t + \frac1{t^2}\right)\,dt} = e^{-\ln t + \frac1t} = \dfrac1t e^{1/t}\)

This means 

\((y\mathcal I)' = \left(1 + \dfrac1t\right)^2 \mathcal I\)

\(\dfrac{y}t e^{1/t} = \displaystyle \int \dfrac1t \left(1 + \dfrac1t\right)^2 e^{1/t}\,dt\)

You will need some special functions to evaluate this integral, so good luck. Hope this helps.

No, (5, 5) is also possible.

\(\angle APC = 115^\circ\\ \angle BPC = 65^\circ\\ \because CP = CB \\ \therefore \angle CBP = 65^\circ\\ \because AB = AC\\ \angle ACB = 65^\circ\\ \)

Then you consider △CPB to find angle PCB. The rest is trivial.

Even integers are multiples of 2, which means they are 2 times "something".

5 times 2 times "something" is 10 times "something". That means "5 times an even integer" is the multiple of 10.

The multiples of 10 (less than 500) are 10, 20, 30, 40, ..., 490.

Therefore there are 49 such integers.

Recall the definition of a^b.

\(a^b = \underbrace{a\times a \times a \times\cdots\times a}_{b\text{ times}}\)

That means 

\(10^2 = 10\times 10\\ 4^3 = 4\times 4 \times 4\)

The rest is simple.

Let a, b, c be the edges.

\(\begin{cases}ab = 3\\bc = 2\sqrt 3\\ca = \sqrt{15}\end{cases}\)

(1) * (3) / (2) : 

\(a^2 = \dfrac{3\sqrt{15}}{2\sqrt 3} = \dfrac{3\sqrt 5}2\\ a = \sqrt{\dfrac{3\sqrt 5}2}\)

Similarly, 

\(b = \sqrt{\dfrac6{5}\sqrt5}\) and \(c = \sqrt{2\sqrt 5}\)

Required ratio = b : c = \(\sqrt 3 : \sqrt 5\)

(y - 5)^2 = (x - 5)^2

|y - 5| = |x - 5|

y - 5 = x - 5 or y - 5 = 5 - x

y = x or y = 10 - x

The area of a regular hexagon A is related to its length L by 

\(A = \dfrac{3\sqrt 3}2 L^2\)

Consider the top white isosceles triangle.

The base is \(2L\sin 60^\circ = \sqrt 3 L\). This is also the length of the rectangle.

The width of the rectangle is \(L\), so the area of the rectangle is \(\sqrt 3 L^2 = \dfrac23 A\).

The area of rectangle is two thirds of the area of the hexagon, which is 60.

The equation implies that (a - 1) is a multiple of 13 and (b - 1) is a multiple of 11. 

The smallest possible a + b occurs when a - 1 = 13 and b - 1 = 11, which means min(a + b) = 25.