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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.

Did you know that AoPS problems are not allowed on this website?

It is because AoPS is called Art of Problem Solving, not Art of Asking Others to Do Problems for you.

If you seriously need help, I believe there is support on their site.

You can still answer it, it does not matter that I have answered already.

Consider the cases where you pick 0, 1, 2, 3, 4 cherry candies respectively. You can calculate how many candies you can pick after picking some amount of cherry candies, and suppose you can pick k more candies after picking some cherry candies, these are the only cases:

So if you are to pick k more candies from apple candies and banana candies, you would have k + 1 cases (just set k to some small number and count the number of rows in the table above). 

Now you should be able to manually count the cases.

For the domain, \(\text{denominator} \neq 0\). Then we have \(\lfloor x^2 - 17x + 18\rfloor \neq 0\).

By properties of floor function, we have \(x^2 - 17x + 18 < 0\text{ or }x^2 - 17x + 18 \geq 1\).

Now you can solve the compound inequality to get the domain of f(x).

Note that it is not possible to divide evenly into 144 but not 72, since 144 = 72 • 2.

Factors of 196: 1, 2, 7, 14, 28, 98, 196

We see that 7, 14, 28, 98, and 196 divide evenly into 196, but not 72, since 72 is not divisible by 7.

The answer is obvious now.

4 out of 32 are canines, so 28 out of 32 are not canines.

Percentage = \(\dfrac{28}{32} \times 100\%\), which you should be able to compute with a calculator.

Remember that you get the percentage by multiplying the proportion by 100%.

First, to ensure that each box must have one of each type, Marvin could place one of each type into each box before he starts randomizing the contents.

After he does so, there are 3 Scuba Steve figures and 2 Diving Dan figures left. Now he can place the figures into the boxes as usual, ignoring the constraints.

Is this easier now?

Let's see what would happen if we write the problem in mathematical expression.

\((2y + 1)(x - 4) = xy + 1\\ 2xy + x - 8y - 4 = xy + 1\\ xy + x - 8y - 5 = 0\)

This is still very hard to manipulate, but let us do it like this:

\(xy + x - 8y - 8 + 3 = 0\\ xy + x - 8y - 8 = -3\\ x(y + 1) - 8(y + 1) = -3\\ (x - 8)(y + 1) = -3 \)

Since we get a negative result, we know that x is smaller than 8. We can write it as \((8 - x)(y + 1) = 3\) by multiplying both sides by -1.

Now, since 8 - x and y + 1 are integers, there are only two possibilities: (i) the left-hand side is \(1 \times 3\), (ii) the left-hand side is \(3\times 1\). Which one is 1 and which one is 3? You can solve for the values of x and y respectively in each case, and see if it contradicts the condition "... whole numbers x and y between 1 and 10 inclusive ...".

Please tell me if you are still stuck.

Using the substitution \(u = x + 3\), \(du = dx\), we have the change of integration boundary \(\begin{cases}x = 1 \implies u= 4\\x = 2 \implies u = 5\end{cases}\). Then we have \(\displaystyle\int_1^2 \dfrac{(x + 3)^2}3\,dx =\dfrac13\int_4^5u^2\,du\). You can use the power rule to integrate it. Note that \(\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)\) where F is an antiderivative of f.

Let L, W, H be the length, width, and height of the prism respectively. 

Writing the conditions in mathematical terms,

\(\begin{cases}2(LW + WH + HL) = 56\\4(L + W + H) = 68\end{cases}\)

What we want to find, in mathematical expression, is \(\sqrt{L^2 + W^2 +H^2}\) by Pythagoras theorem.

Recall the identity: \((\color{blue}L+W+H\color{black})^2 = \color{red}(L^2+W^2+H^2)\color{black} + \color{green}2(LW + WH + HL)\)

Does this ring a bell?

Moving some terms yields: \(2x^2 + 2y^2 - x - y - 8 = 0\)

Note that geometrically, this is the equation of a circle, because we can rewrite the equation as follows:

\(\begin{array}{rcl} 2x^2 + 2y^2 - x - y - 8 &=& 0\\ x^2 + y^2 - \dfrac12 x - \dfrac12 y - 4&=&0\\ \left(x - \dfrac14\right)^2 + \left(y - \dfrac14\right)^2 &=& \dfrac{33}8 \end{array}\)

Consider the straight line x - y = c, where c is varying through real numbers.

You can find the maximum value of c such that the line and the circle intersects, i.e., \(\begin{cases}2x^2 + 2y^2 - x - y - 8 = 0\\x - y = c​​​​\end{cases}\) has solutions.

To do so, you need to substitute y = x - c into the circle equation, then consider the discriminant of the resulting equation. Since the system has solutions, the discriminant of the resulting equation must be nonnegative. Solving the equation \(\Delta \geq 0\) will give you the suitable range of values of c.

Please tell me if you want more explanations or you need more help.