Let AB = b, and AC = c. Moreover, let \([ABX] = A_1\) and \([ACX] = A_2\), where \([\cdot]\) denote "area of".

By Stewart's theorem, we have 

\((13 + 5)(13^2 + 13(5)) = 5b^2 + 13c^2\\ 5b^2 + 13c^2 = 4212 \)

By basic geometry, we have \(\dfrac{A_1}{A_2} = \dfrac{13}5\). 

Recall that the circumradius of a triangle with sides a, b, c and area A is \(\dfrac{abc}{4A}\). Then, since the circumradius of triangles ABX and ACX are the same, we have 

\(\dfrac{13^2 b}{4A_1} = \dfrac{5(13)c}{4A_2}\\ \dfrac{A_1}{A_2} = \dfrac{13b}{5c}\\ \dfrac{13}5 = \dfrac{13b}{5c}\\ b = c\).

Hence, 

\(18b^2 = 18c^2 = 4212\\ b = c = 3 \sqrt{26}\)

We calculate the half perimeter of triangle ABC to be \(\dfrac{18 + b + c}2 = 3 \sqrt{26} + 9\). Then, by Heron's formula, the area required is:

\(\sqrt{(3 \sqrt{26} + 9)(3\sqrt{26} - 9)(9^2)} = 27\sqrt{17}\). 

\(m_{BC } = \dfrac{2 - (-8)}{-3 - 1} = -\dfrac52\\ m_{\text{altitude}} = \dfrac25\\ \dfrac{y - 3}{0 - 2} = \dfrac25\\ y - 3 = -\dfrac45\\ y = \dfrac{11}5\)

\(1000 \left(-\dfrac89\right)^{n-1} > 1\)

To be greater than 1, it must be greater than 0, so we only count the odd numbered terms, i.e. 1st term, 3rd term, etc.

After some point, even if it is an odd numbered term, it won't be greater than 1 anymore.

Since the term is an odd numbered term as we deduced, we can safely let n = 2k + 1 for k >= 0, k is an integer.

\(1000 \left(\dfrac{64}{81}\right)^k > 1\\ k < \log_{64/81}{\dfrac1{1000}}\\ k < 29.3\)

Then any nonnegative integer k with k < 29 will correspond to a positive term in the original sequence which is greater than 1.

Hence, the answer is 29.

We will tackle the general case first. What is the sum of all positive integers less than 1000 ending in a digit n?

The numbers must be of the form \(10m + n\) where \(0 \leq m \leq 99\).Therefore, the answer will be \(\displaystyle \sum_{m = 0}^{99} (10m + n) = 100n + 49500\).

Hence, the answer to your original question will be \(4(49500) + 100(3 + 4 + 6 + 9) = 200200\).

Simply divide by the amount you have overcounted using 7!.

\(\text{Answer} = \dfrac{7!}{3!2!2!} = 210\)

\(\quad\text{Required probability}\\ = \dfrac35 \cdot \dfrac24 \cdot \dfrac23 + \dfrac25 \cdot \dfrac34 \cdot \dfrac13\\ =\dfrac3{10}\)

Suppose the friends are A, B, C, Dhruv (D), E. Then denote the number of chocolate gotten by each person as \(x_A, x_B, \cdots, x_E\).

The problem becomes to count the nonnegative integer solutions of \(x_A+x_B+x_C+x_D+x_E=8\) with \(x_D \geq 6\).

We make the substitution \(y_A=x_A, y_B=x_B, y_C=x_C, y_D=x_D-6,y_E=x_E\). Then the equation becomes \(y_A+y_B+y_C+y_D+y_E=2\). We will count the nonnegative integer solutions of this equation to get the answer. The answer is \(\displaystyle \binom{2 + 5 - 1}{5 - 1} = 15\).

The pair \((x_2, x_5)\) can be (0, 1), (0, 3), ..., (0, 9), (2, 1), (2, 3), ..., (2, 7), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (8, 1).

From here, 5 pairs add up to 9, 4 pairs add up to 7, 3 pairs add up to 5, 2 pairs add up to 3, and 1 pair add up to 1.

If \(x_2 + x_5 = k\), then \(x_1 + x_3 + x_4 + x_6 + x_7 = 10 - k\). The number of nonnegative integer solutions of \(x_1 + x_3 + x_4 + x_6 + x_7 = N\) is \(\displaystyle \binom{N+5-1}{5-1} = \binom{N+4}4\). Then, we simply calculate the required number to be \(\displaystyle 5\binom{1 + 4}{4} + 4\binom{3 + 4}{4} + 3\binom{5 + 4}{4} + 2\binom{7+ 4}{4} + 1\binom{9+4}{4} = 1918\).

\(\begin{cases} x + y = 10\\ x^2 + y^2 = 62 + 2xy \end{cases}\\ (x - y)^2 = 62\\ x - y = \pm \sqrt{62}\\ \)

When x - y = \(\sqrt{62}\), \((x, y) = \left(\dfrac{10 + \sqrt{62}}2, \dfrac{10 - \sqrt{62}}2\right)\).

Otherwise, \((x, y) = \left(\dfrac{10 -\sqrt{62}}2, \dfrac{10 + \sqrt{62}}2\right)\).

These 2 pairs are all the solutions.

Suppose the answer is x%.

\(60\% \cdot 30\% + (1 - 60\%) x\% = 50\%\\ 0.18 + 0.004x = 0.5\\ x = 80\)

80% of the remainder of the games must be won.

The polynomial is just x^2 - 24x + c. When it is the square of a binomial,

\(24^2 - 4(1)(c) = 0\\ c = 144\)

The polynomial is just x^2 + sx + 4. For the polynomial to be the square of a binomial, the discriminant is equal to 0.

\(s^2 - 4(1)(4) = 0\\ s^2 = 16\\ s = \pm 4\)

\(\displaystyle \frac{x}{x - 5} = \frac{4}{x - 4} + \frac{3}{x} - 2\\ \dfrac x{x-5} = \dfrac{7x-12-2x(x-4)}{x(x-4)} = \dfrac{-2x^2+15x-12}{x^2 - 4x}\\ x^3 - 4x^2 = (x - 5)(-2x^2 + 15x - 12)\\ x^3 - 4x^2 = -2x^3 + 25x^2 - 87x + 60\\ 3x^3 - 29x^2 + 87x - 60 = 0 \)

Solving numerically (or using Cardano's formula or Tartaglia's formula) we have \(x \approx 0.97404\text{ or }x \approx 4.3463 \pm 1.2816i\)

By Vieta's formulae we have b = -(5 + 3i + (8-2i)) = -13 - i and c = (5 + 3i)(8 - 2i) = 46 + 14i so b + c = 33 + 13i.

Case 1: \(c \leq -5\)

Then 

\(-c-5-3c = 10 - 2c+8+6c\\ -8c - 23 = 0\\ c = -\dfrac{23}8\)

But then c is not less than or equal to -5. This root is rejected.

Case 2: \(-5 < c \leq 0\)

Then

\(c + 5 - 3c = 10 - 2c + 8 +6c\\ -6c - 13 = 0\\ c = -\dfrac{13}6\)

This root is in the range -5 < c <= 0. So we keep it.

Case 3: \(0 < c \leq 4\)

Then

\(c + 5 - 3c = 10 - 2c + 8 -6c \\ 6c = 13\\ c = \dfrac{13}6\)

This root is in range 0 < c <= 4. So we keep it.

Case 4: \(c > 4\)

Then

\(c + 5 - 3c = 10 + 2c - 8 - 6c\\ 2c + 3 = 0\\ c = -\dfrac32\)

But then c is not greater than 4. This root is rejected.

Hence, the solutions are: \(c = \dfrac{13}6\), \(c = -\dfrac{13}6\).