MaxWong

It looks something like this.

We try to pick out all the combinations of terms, one from each polynomial, such that their product is some number times \(x^2\).

First, if we pick 2x^3 from the first polynomial, it is impossible to get x^2 by multiplying it to another term from the other polynomial, we do not care about this case in this scenario.

Then, we consider the second term x^2 from the first polynomial. To get some multiple of x^2, we can only choose the constant term from the second polynomial, which will give us \(+15x^2\) if we multiply the terms.

Then, for -3x, we choose -11x from the second polynomial, so when we multiply them, we get \(+33x^2\)

For +5, we choose tx^2 from the second polynomial, so when we multiply them we get \(+5tx^2\).

Therefore, the term in x^2 is \(15x^2 + 33x^2 + 5tx^2\), which is \((5t + 48)x^2\). Since we are given that the product has no x^2 term, the coefficient must be 0, which gives \(t= -\dfrac{48}5\).

Considering the sum of interior angles of the triangle we have \(\angle B = 34^\circ\). Then we have, by sine rule,

\(\dfrac{5}{\sin 34^\circ} = \dfrac{\overline{BC}}{\sin 108^\circ}\)

From here, we have \(\overline{BC} \approx8.50\).

Simplifying, the quadratic equation is \(x^2 + 20x - 13 = 0\). Therefore, a + b = -20 and ab = -13.

Now, 

\(\quad (a^2 + a + 1) + (b^2 + b + 1)\\ = (a^2 + b^2) + (a + b) + 2\\ = (a + b)^2 - 2ab + (a + b) + 2\\ = 400 + 26 - 20 + 2\\ = 408\)

Also,

\(\quad (a^2 + a + 1)(b^2 + b + 1)\\ = \dfrac{(a^3 -1)(b^3 - 1)}{(a - 1)(b - 1)}\\ = \dfrac{(ab)^3 - (a^3 + b^3) + 1}{ab - (a + b) + 1}\\ = \dfrac{(ab)^3 - (a + b)^3 + 3ab(a+b) + 1}{ab - (a + b) + 1}\\ = 823\)

Hence, the required equation is \(x^2 - 408x + 823 = 0\).

Simplifying, the equation you have is \(x^2 - 23x - 16 = 0\).

From Vieta's formulas, we have \(\begin{cases} a + b = 23\\ ab = -16 \end{cases}\).

Now, note that \(a^4 + b^4 = (a^2 + b^2)^2 - 2a^2 b^2 = ((a+b)^2 -2ab)^2 - 2(ab)^2\).

Can you continue from here?

p + r = 9 + p means r = 9.

From q - 2r = -2 + q, we have r = 1.

These two equations are contradictory. Hence, the system has no solutions.

Notations: We use the following notation \(\displaystyle \binom{n}k\) to mean \(\displaystyle \binom{n}k = C^n_k\).

Note that \(\displaystyle \binom{20}{10} = 184756\) so all possible paths are exhausted by the penguins.

There is an important observation in all 4 problems, which I will state below:

Observation:

Let \((x, y)\) be a lattice point, with \(0 \leq x \leq 10\), \(0 \leq y \leq 10\).

Each path that passes through this lattice point must start with a sequence of x + y moves,

consisting of x moves to the right, and y moves upwards, in some order.

By observation, to count the number of path passing through (x, y), we are counting the number of permutations of x (indistinguishable) right moves and y (indistinguishable) up moves. That number is \(\displaystyle \binom{x + y}{x}\). From there, we move from (x, y) to (10, 10) by 10 - x right moves and 10 - y up moves. By the same logic, there are \(\displaystyle \binom{(10 - x) + (10 - y)}{10 - x} = \binom{20 - x - y}{10 - x}\) ways to move from (x, y) to (10, 10). 

Therefore, by basic principles of combinatorics, there are \(\displaystyle \binom{x + y}x \cdot \binom{20 - x - y}{10 - x}\) different paths that passes through \((x, y)\).

With this in hand, the problems are easily solved.

Problem 1

The total sum is \(\displaystyle \binom{7 + 10}{7} \cdot \binom{20 - 7 - 10}{10 - 7} + \binom{8 + 9}8 \cdot \binom{20 - 8 - 9}{10 - 8} + \binom{9 + 8}{9} \cdot \binom{20 -9-8}{10-9} + \binom{10 + 7}{10} \cdot \binom{20 -10- 7}{10-10}\), which is 184756. 

Seeing this answer, you may think there is an easier way to do this. In fact, there is. Note that for each penguin, after 17 moves, one of the 4 points must be reached. From each of the 4 points, the other 3 points cannot be reached. Therefore, each penguin passes through one and only one penguin counter. Therefore the answer is exactly the number of penguins we have.

Problem 2

From the observation in Problem 1, if you fix a number k such that \(0 \leq k \leq 20\) and place counters on each point (x, y) such that \(x + y = k\), every penguin will be counted exactly once. (In Problem 1, the case k = 17 is shown. )

The problem then becomes "how many ways are there to choose 2 integers from 0 to 20 inclusively", which is \(\displaystyle \binom{21}2 = 210\).

Problem 3

The expected sum is 11 times the average number of paths passing through a point, i.e., in summation notation, \(11 \cdot \dfrac{\displaystyle \sum_{x = 0}^{10} \sum_{y = 0}^{10} \binom{x + y}x \binom{20 -x-y}{10-x}}{11^2}\)

While it may seem intimidating, there is a clever way to calculate the double summation \(\displaystyle \sum_{x = 0}^{10} \sum_{y = 0}^{10} \binom{x + y}x \binom{20 -x-y}{10-x}\). From the observation in Problem 2, the sum of counters over the lattices points with \(x + y = k\) is just the total number of penguins, where \(0 \leq k \leq 20\) is a fixed constant. Therefore, the sum is equal to \(\displaystyle\sum_{k = 0}^{20} 184756 = 21(184756) = 3879876\). Now we substitute into the formula and get the expected sum to be \(\dfrac{3879876}{11} = \boxed{352716}\).

Problem 4

The answer is just \(\displaystyle \sum_{x = 4}^{4} \sum_{y = 0}^{10} \binom{x + y}x \binom{20 -x-y}{10-x} = \sum_{y = 0}^{10} \binom{4 + y}4 \binom{16 - y}{6}\). I will let you figure out the rest.

I guess you are talking about:

\(T_f = \{f^{2021}(s): s \in S\}\), find the remainder when \(\displaystyle \sum_{f\in F} |T_f|\) is divided by 2017, am I correct?