MaxWong

Alternative method:

We know that 

0.333... = 1/3

Multiplying both sides by 3:

0.999... = 1

0.999... = 0.9 + 0.09 + 0.009 + 0.0009 + ... = \(9 \displaystyle\sum_{n = 1}^\infty \dfrac{1}{10^n} \)

By sum of infinite geometric series formula, 0.999... = \(9 \cdot \dfrac{\dfrac1{10}}{1-\dfrac{1}{10}} = 9\cdot \dfrac19 = 1\)

That means \(x^2 + bx + c \leqslant 0\) when \(x\in \left[-2, 3\right]\).

And this also means \(x^2 + bx + c < 0\) when \(x\in \left(-2, 3\right)\).

When \(x^2 + bx + c = 0\), \(x = -2\text{ or }x = 3 \)

So \(x^2 + bx + c = (x + 2)(x - 3) = x^2 - x - 6\).

Comparing coefficients, b = -1 and c = -6.

b + c = -1 + (-6) = -7.

This is seemingly easy, but it is extremely hard.

This one is the Collatz conjecture, which remains unsolved for years. 

Link: https://en.wikipedia.org/wiki/Collatz_conjecture

EDIT: Sorry, I read the question wrong. But it is still good to read about the conjecture.

f(11) = 34, so f(34) is defined.

f(34) = 34/2 = 17, so f(17) is defined.

f(17) = 3(17) + 1 = 52, so f(52) is defined.

f(52) = 52/2 = 26, so f(26) is defined,

f(26) = 26/2 = 13, so f(13) is defined.

f(13) = 3(13) + 1 = 40, so f(40) is defined.

f(40) = 40/2 = 20, so f(20) is defined.

f(20) = 20/2 = 10, so f(10) is defined.

f(10) = 10/2 = 5, so f(5) is defined.

f(5) = 3(5) + 1 = 16, so f(16) is defined.

f(16) = 16/2 = 8, so f(8) is defined.

f(8) = 8/2 = 4, so f(4) is defined.

f(4) = 4/2 = 2, so f(2) is defined.

f(2) = 2/2 = 1, so f(1) is defined.

f(1) = 3(1) + 1 = 4

Now we ended up in an infinite loop of 4, 2, 1, 4, 2, 1, etc.

Let's count how many numbers did we encounter.

For \(x\in\{11,34,17,52,26,13,40,20,10,5,16,8,4,2,1\}\), f(x) is defined.

Therefore there are at least 15 integers in the domain of f.

This means

\(n^2 < 3^5\\ n^2 < 243\)

Because 15² = 225 < 243, and 16² = 256 > 243,

The largest n such that n²⁰⁰ < 3⁵⁰⁰ is n = 15.

https://web2.0calc.com/questions/please-help_16171#r1

When x <= 3,

f(x) = 0

2x + 1 = 0

x = -1/2

When x > 3,

f(x) = 0

8 - 4x = 0

x = 2 

but x has to be greater than 3, so this is not actually a solution.

The sum of all values of x such that f(x) = 0 is -1/2.

If you and your dad do not understand any part of my solution, feel free to ask me here. :)

Wish you happy problem-solving! :D

We express x^2 + y^2 and x^3 + y^3 in terms of (x + y) and (xy).

\(x^2 + y^2 = (x + y)^2 - 2(xy)\\ x^3 + y^3 = (x + y)((x + y)^2 - 3(xy))\)

Now, let x + y = t, xy = u.

\(\begin{cases}t^2 - 2u = 7\\t(t^2 - 3u)=10\end{cases}\)

Substituting the first equation into the second,

\(\begin{cases}t^2 - 2u = 7\\t(7 - u)=10\end{cases}\)

Manipulating the second equation, \(u = 7 - \dfrac{10}t\)

Substituting, 

\(t^2 - 14 + \dfrac{20}t = 7\\ t^3 - 21t + 20 = 0\)

By factor theorem, (t - 1) is a factor of the left hand side.

\((t - 1)(t^2 + t - 20) = 0\\ (t - 1)(t - 4)(t + 5) = 0\\ t = 1 \text{ or }t = 4 \text{ or } t= -5\)

Therefore, the possible values of (x + y) are 1, 4, and -5. 

If Fred starts on 1,

There's a 1/2 probability that it moves to 0 directly, and a 1/2 probability that it moves to 2.

Once it moves to 2, it must move back to 1.

And when it moves back to 1 again, there are again 1/2 probability that it moves to 0, and 1/2 probability that it moves to 2.

This repeats again and again.

So that means there are 1/2 probability it reaches 0 in 1 step, 1/4 probability it reaches 0 in 3 steps, 1/8 probability it reaches 0 in 5 steps, etc.

Because the e_1 = sum of all possible (probability * steps),

\(e_1 = \dfrac12 + \dfrac34 + \dfrac58 + \dfrac7{16} + \cdots = \displaystyle\sum_{n = 1}^\infty \dfrac{2n -1}{2^n}\) --- (1)

Now, we consider (e_1)/2. (The standard way of evaluating an A.G.S. (arithmetico-geometric series))

\(e_1 \cdot \dfrac12 = \dfrac14 + \dfrac38 + \dfrac5{16}+\cdots\) --- (2)

Subtracting (2) from (1),

\(\dfrac{e_1}2 = \dfrac12 + \left(\dfrac{2}4 + \dfrac28 + \dfrac2{16}+\cdots\right) = 1 + \left(\dfrac14 + \dfrac18 + \cdots\right) = \dfrac32\)

\(e_1 = 3\)

Therefore, if Fred starts at 1, then it is expected to reach 0 in 3 steps.

By a similar idea as above, we can get \(e_2 = 2\cdot \dfrac12 + 4\cdot \dfrac14 + 6\cdot \dfrac18 + \cdots = \displaystyle \sum_{n = 0}^\infty \dfrac{n + 1}{2^n} \)

By a similar approach, we can evaluate e_2 to be 4.

Therefore, (e_1, e_2) = (3, 4).