MaxWong

Hint: This looks complicated, but you can rewrite it into the system of inequalities \(\begin{cases}\dfrac t3 - 5 < t - 2 + 2t\\t - 2 + 2t \leq -3t + 7\end{cases}\), solve each inequality, and find the range of values of t which satisfies both inequalities. Now it just becomes two separate linear inequalities in t.

For l(x) to make sense, it is required that \(\dfrac{x - 1}{x - 3} \in (-1, 1)\).

Then, we solve the inequality \(-1 < \dfrac{x - 1}{x - 3} < 1\) to find the domain of f.

I will do one side of the inequality as an example. Let's say we want to solve \(\dfrac{x - 1}{x - 3} < 1\). We can't just multiply (x - 3) on both sides because it is not guaranteed that x - 3 is positive. Instead, we multiply (x - 3)^2 on both sides since it is always nonnegative. Also, note that x cannot be 3 because of the denominator.

\(\dfrac{x - 1}{x - 3} \cdot (x - 3)^2 \leq (x - 3)^2, \,x \neq 3\\ (x - 1)(x - 3) \leq (x - 3)^2, \, x \neq 3\\ x^2 - 4x + 3 \leq x^2 - 6x + 9,\, x \neq 3\\ 2x - 6 \leq 0, \, x \neq 3\\ 2x - 6 < 0\\ x < 3 \)

Now, you can solve the other inequality \(-1 < \dfrac{x - 1}{x -3}\) to get another inequality for x, combine them, and you get the domain of \(l(x)\).

There are infinitely many pairs of solutions for (x, y).

If two quadratic polynomials have the exact same roots, then one of them is a multiple of the other.

Judging by the coefficients of x, Ax^2 + x + B should be 1/8 times of x^2 + 8x + 4.

You can continue by expanding \(\dfrac18 (x^2 + 8x + 4)\) and finding the coefficient of x^2 and the constant term.

Since it is a translation, \(AA'B'B\) is a parallelogram or a straight line.

The diagram is shown below:

From the diagram, you can see that BA' = 12.

Hint: If a number \(N\) in base 10 has 4 digits when written in base 3, that means \(27 \leq N < 81\), because 27 is the least number to have 4 digits in base 3 (which is 1000), and 81 is the least number to have 5 digits in base 3 (which is 10000).

Also, if it has 3 digits when written in base 5, \(25 \leq N < 125\) for a similar reason.

If an integer satisfies both inequalities, that means \(27 \leq N < 81\).

I believe you can continue on your own.

But in case you are still clueless, feel free to tell me.

Let \( d = \deg (p) \in \mathbb Z^+\).

Note that \(\deg(p\circ p) = d^2\) and \(\deg(x\cdot p + x^2 + 3x + 1) = \max(d + 1, 2)\)

Now we eliminate the case when p(x) is constant to ensure that \(\max(d + 1, 2) = d + 1\).

(Because if p(x) = c, c = cx + x^2 + 3x + 1 for any x, which is impossible.)

Comparing the degrees on both sides gives \(d^2 = d + 1\), which has no solution in \(\mathbb Z^+\).

Therefore, there are no such polynomial p(x) with p(p(x)) = xp(x) + x^2 + 3x + 1.

Note that \(\varphi(13) = 13 - 1 = 12\), where \(\varphi\) denotes the Euler totient function.

Now, for these type of problems, you can take mod 13 of the base and mod \(\varphi(13)\) of the power.

\(\begin{array}{rcl} 333^{333} &\equiv& 8^9 \pmod{13}\\ &=& 2^{27} \pmod{13}\\ &=& 2^3 \pmod{13} \end{array}\)

Note that I used \(8^9 = (2^3)^9 = 2^{27}\) and then took mod \(\varphi(13)\) of the power again.

Now the size of the problem becomes manageable, because you can calculate 2^3, and that's the answer.

Yes, that's true. I checked with Wolfram Alpha.

You can pretty much learn on your own using some online resources.