heureka

It should work like this:
\(f(5, 3) = 555 \\ f(12,3)=121212 \\ \\ 555=5\cdot10^2+5\cdot10^1+5\cdot10^0 \\ 121212=12\cdot10^4+12\cdot10^2+12\cdot10^0 \)

We see that the power of ten dependence on the length of the number that we're repeating.

Knowing this we can bring out the formula.

\(rep\left(x,n\right)=\sum_{c=0}^{n-1}x\cdot10^{c\lfloor\log_{10}x\rfloor}\)

Where  (x is the repeating number)

Where  (n is the number of repetitions)

\(\lfloor log_{10}x\rfloor\) - length of x

Sorry there is a mistake:

The length of \(x\) is \(\mathbf{\lfloor log_{10}x\rfloor +\color{red} 1 }\)

\(\large{rep\left(x,n\right)=\sum \limits_{c=0}^{n-1}x\cdot10^{c \left( \lfloor\log_{10}x\rfloor+1 \right) }}\)

see Desmos:

hint:

\(\begin{array}{|rcll|} \hline &&\mathbf{ \lfloor\log_{10}x\rfloor+1 } \\ &=& \lfloor\log_{10}x+1 \rfloor \\ &=& \lfloor\log_{10}x+\log_{10}(10) \rfloor \\ &=&\mathbf{ \lfloor\log_{10}(10x) \rfloor } \\ \hline \end{array}\)

see Desmos:

Strict ranges (number can't be more/less than...)

I assume:

\(\begin{array}{|rcll|} \hline l(s,x) &=& \max(s,x) \\ g(s,x) &=& \min(s,x) \\ \hline lg(m,x,i) &=& g(m, l(i, x)) \\ \mathbf{lg(m,x,i)}&=& \mathbf{ \min(m,\max(i,x)) } \\ \hline \end{array}\)

hint: \(\boxed{\max(s,x) - \min(s,x) = |s-x|}\)

In triangle \(\triangle ABC\), a point \(D\) is on \(\overline{AC}\) so that \(AB = AD\) and \(\angle ABC - \angle ACB = 32^{\circ}\).
Find \(\angle DBC\) in degrees.

\(\begin{array}{|rcll|} \hline \angle ABD &=& \angle ADB \quad & | \quad \angle ADB = \angle DBC + \angle ACB \\ \angle ABD &=& \angle DBC + \angle ACB \quad & | \quad \angle ABD = \angle ABC - \angle DBC \\ \angle ABC - \angle DBC &=& \angle DBC + \angle ACB \\ 2\angle DBC &=& \angle ABC - \angle ACB \quad & | \quad \angle ABC - \angle ACB = 32^{\circ} \\ 2\angle DBC &=& 32^{\circ} \\ \mathbf{\angle DBC} &=& \mathbf{16^{\circ}} \\ \hline \end{array} \)

Triangle \(ABC\) is acute. Point \(D\) lies on \(\overline {AC}\) so that \(\overline {BD}\perp \overline {AC}\),
and point \(E\) lies on \(\overline {AB}\) such that \(\overline {CE}\perp\overline {AB}\).
The intersection of segments \(\overline {CE}\) and \(\overline {BD} \)is \(P\).
Find the value of \(\overline{AE}\) if \(\overline{CP}=10\), \(\overline{PE}=20\), and \(\overline{EB}=30\).

Thank you, Melody !

Why 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?

Hello CalculatorUser

No, it is because that guaruntees  that after Martin and Ron the sum of removed marbles is 5.

or in other words Ron supplements each time to 5.

Here is the course

\(\begin{array}{|r|r|l|} \hline \text{Martin} & \text{Ron} & \text{sum} \\ \hline & 3 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ 1 \\ \hline \end{array}\)

Ron and Martin are playing a game with a bowl containing 39 marbles.
Each player takes turns removing 1, 2, 3 or 4 marbles from the bowl.
The person who removes the last marble loses.
If Ron takes the first turn to start the game,
how many marbles should he remove to guarantee he is the winner?

Ron first takes 3 marbles.

Than, if Martin takes x marbles, Ron takes 5-x marbles, this repeats 7 times.

The last marble is then for Martin.

39 = 3 + 7*5 + 1

At Pete's Pizza Palace there are four different toppings you can get on a pizza

(all pizzas come with cheese and sauce):
Onions, Strawberries, Pepperoni, and Chocolate.
How many different pizzas can you order at Pete's?
You can have any combination of the toppings you want including no toppings or all four toppings.

\(\begin{array}{|c|cccc|} \hline \text{binary} & \text{Onions} & \text{Strawberries} & \text{Pepperoni} & \text{Chocolate} \\ \hline 0 & 0 & 0& 0& 0 & \\ \hline 1 & 0 & 0& 0& 1 & \\ \hline 2 & 0 & 0& 1& 0 & \\ \hline 3 & 0 & 0& 1& 1 & \\ \hline 4 & 0 & 1& 0& 0 & \\ \hline 5 & 0 & 1& 0& 1 & \\ \hline 6 & 0 & 1& 1& 0 & \\ \hline 7 & 0 & 1& 1& 1 & \\ \hline 8 & 1 & 0& 0& 0 & \\ \hline 9 & 1 & 0& 0& 1 & \\ \hline 10& 1 & 0& 1& 0 & \\ \hline 11& 1 & 0& 1& 1 & \\ \hline 12& 1 & 1& 0& 0 & \\ \hline 13& 1 & 1& 0& 1 & \\ \hline 14& 1 & 1& 1& 0 & \\ \hline 15& 1 & 1& 1& 1 & \\ \hline \end{array}\)

There are 16 different pizzas

For a certain value of \(k\), the system
\(\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}\)
has no solutions. What is this value of  \(k\)?

Find all values of \(k\) such that \(3\cot(4k-\pi)=\sqrt3\)

\(\begin{array}{|rcll|} \hline \mathbf{3\cot(4k-\pi) } &=& \mathbf{ \sqrt3 } \quad &| \quad \cot(4k-\pi) = \dfrac{1}{\tan(4k-\pi)} \\\\ \dfrac{3}{\tan(4k-\pi)} &=& \sqrt3 \quad &| \quad \cdot \tan(4k-\pi) \\\\ 3 &=& \sqrt3 \cdot \tan(4k-\pi) \\\\ \sqrt3 \cdot \tan(4k-\pi) &=& 3 \quad &| \quad \cdot \sqrt3 \\ 3 \cdot \tan(4k-\pi) &=& 3\sqrt3 \quad &| \quad : 3 \\ \tan(4k-\pi) &=& \sqrt3 \\ \tan\Big( -(\pi-4k) \Big) &=& \sqrt3 \quad &| \quad \tan(-x)= - \tan(x) \\ -\tan(\pi-4k) &=& \sqrt3 \quad &| \quad \tan(\pi-x) = -\tan(x) \\ - \Big(-\tan(4k) \Big) &=& \sqrt3 \\ \mathbf{\tan(4k)} &=& \mathbf{\sqrt3} \quad & | \quad \arctan() \\ \arctan\Big(\tan(4k)\Big) &=& \arctan(\sqrt3) + n\cdot \pi \qquad n \in \mathbb{Z} \\ 4k &=& \arctan(\sqrt3) + n\cdot \pi \quad & | \quad \arctan(\sqrt3)=\dfrac{\pi}{3} \\ \\ 4k &=&\dfrac{\pi}{3}+ n\cdot \pi \quad & | \quad : 4 \\ \\ \mathbf{ k } &=& \mathbf{\dfrac{\pi}{12}+ n\cdot \dfrac{\pi}{4} } \qquad n \in \mathbb{Z} \\ \hline \end{array}\)