Because the tank doubles everr minute, the tank is half full in 9 minutes.

Note that from \(3^x=\left(5^w+2^y\right)\left(5^w-2^y\right)\) we can infer \(5^w+2^y=3^x,5^w-2^y=1\) because if \(5^x+2^y=3^m,5^x-2^y=3^{x-m}\) then \(\left(5^w+2^y\right)+\left(5^w-2^y\right)=2.5^w=3^m+3^{x-m}⋮3\) (a contradiction)

We will show there is exactly one set of solution, namely x = y = z = 2. To simplify the equation, we consider modulo 3. We have:

   \(1=0+1^y\equiv3^x+4^y=5^z\equiv\left(-1\right)^z\left(mod3\right)\)

It follows that \(z\) must be even, say z = 2w, then:

 \(3^x=5^z-4^y=5^{2w}-2^{2y}=\left(5^w+2^y\right)\left(5^w-2^y\right)\)

\(\Rightarrow\left\{{}\begin{matrix}5^w+2^y=3^x\\5^w-2^y=1\end{matrix}\right.\)

​ ​​ ​\(\Rightarrow\left\{{}\begin{matrix}\left(-1\right)^w+\left(-1\right)^y\equiv0\left(mod3\right)\\\left(-1\right)^w-\left(-1\right)^y\equiv1\left(mod3\right)\end{matrix}\right.\)

=> \(w\) is odd and \(y\) is even. If \(y>2\) then \(5\equiv5^w+2^y=3^x\equiv1\)or \(3\) (mod 8), a contradiction. So \(y=2\). Then \(5^w-2^y=1\) => \(w=1,z=2\)  and finally \(x=2\).

We have:

\(\left(x+y+z\right)^3-\left(x^3+y^3+z^3\right)=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Rightarrow8=\left(3-z\right)\left(3-x\right)\left(3-y\right)\)   (1)

Since \(\left(3-x\right)+\left(3-y\right)+\left(3-z\right)=9-\left(x+y+z\right)=9-3=6\)  (2)

From (1) and (2) we infer \(\left(x,y,z\right)\) would be (1,1,1), (-5,4,4), (4,-5,4), (4,4,-5).

Without loss of generality, we may assume gcd(a,b,c) = 1.

(otherwise, if d=gcd(a,b,c) the for a'=a/d, b'=b/d, c'=c/d, the equation still holds for a', b', c' and a'b'c' is still a cube if only if abc is a cube).

We multiply equation by abc, we have:

  \(a^2c+b^2a+c^2b=3abc\)(*)

if \(abc=\pm1\), the problem is solved.

Otherwise, let p be a prime divisor of abc. Since gcd(a,b,c)=1, the (*) implies that p divides exactly two of a, b,c. By symetry, we may assume p divides a, b but not c. Suppose that the lagest powers of p dividing a, b are m, n, respecively.

If n < 2m, then \(n+1\le2m\)  and \(p^{n+1}\)| \(a^2c,b^2c,3abc\). Hence \(p^{n+1}\)|\(c^2b\), forcing \(\)\(p\)|\(c\) (a contradiction). If n > 2m, then \(n\ge2m+1\) and \(p^{2m+1}\)|\(c^2b,b^2a,3abc\). Hence \(p^{2m+1}\)|\(a^2c\), forcing \(p\)|\(c\) (a contradicton). Therefore n = 2m and \(abc=\Pi p^{3m}\), \(p\)|\(abc\), is a cube.

Since \(3^{\dfrac{1}{3}}>4^{\dfrac{1}{4}}>5^{\dfrac{1}{5}}>...\) we have \(y^z\ge z^y\) if \(y\ge3\).

Since \(1\le x\le y\), the only possible values for (x, y) are (1,1), (1,2), (2,2). These lead to the equations \(1+1=z,1+2^x=z,4+2^z=z^2\).

The firt equation has the unique solution (1,1,2).

The second equarion has no solution because \(2^x>z\).

The third equation has also no solution since \(2^x\ge z^2,z\ge4\) and (2,2,3) is not a solution.

So the equation has unique solution (1,1,2).

a) In quadrilateral DAMB: \(\angle A+\angle D+\angle B+\angle MAB=360^o\) (1)

In triangle MEC: \(\angle E+\angle C+\angle MEC=180^o\)                        (2)

Sum (1) and (2): ​ ​​ ​\(\angle A+\angle B+\angle C+\angle D+\angle E+\angle MAB+\angle MEC=540^o\)

Since \(\angle MEC=\angle MAB\)

We have: \(\angle A+\angle B+\angle C+\angle D+\angle E+360^o=540^o\)

  \(\Rightarrow\angle A+\angle B+\angle C+\angle D+\angle E=540^o-360^o=180^o\)

b) if 2A=B=C=D=E we infer A = 20^o

Let \(Q\left(x\right)=\left(x+1\right)P\left(x\right)-x\),  (*)

since \(P\left(n\right)=\dfrac{n}{n+1}\) we infer:

 \(Q\left(n\right)=\left(n+1\right)P\left(n\right)-n=0\) for \(n=0,1,..,2015\).

Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:

  \(Q\left(x\right)=a\left(x-0\right)\left(x-1\right)...\left(x-2015\right)\)   (**)

In other hand, in (*) we set x = -1 then  \(Q\left(-1\right)=1\). And replace to (**) we have:

  \(1=Q\left(-1\right)=a\left(-1\right)\left(-2\right)...\left(-2016\right)\)

 \(\Rightarrow a=\dfrac{1}{2016!}\)

Finally \(Q\left(x\right)=\dfrac{1}{2016!}\left(x-0\right)\left(x-1\right)...\left(x-2015\right)\)

\(\Rightarrow Q\left(2016\right)=\dfrac{1}{2016!}2016.2015...1=1\)

from (*) \(Q\left(2016\right)=\left(2016+1\right)P\left(2016\right)-2016\)

\(\Rightarrow\left(2016+1\right)P\left(2016\right)-2016=1\)

\(\Rightarrow P\left(2016\right)=\dfrac{2017}{2017}=1\)