Explore Discrete Math Examples

Stuck on this question
Water fills a tank at an increasing rate of 50 litres during the first hour, 75 litres during the second hour, 100 litres during the third hour and so on. The size of the tank is $16\times <!--\; \times \; -->6\times <!--\; \times \; -->3$. All dimensions in metres. i am stuck trying to figure out how long it takes to fill the tank to the nearest hour.

Deducing Inequality from an Equation
I was interested in Solution of this Non-Homogenous Recurrence Relation
$f(n)=f(n-<!--\; -\; -->1)+f(n-<!--\; -\; -->3)+1$
The Base conditions are:
$f(0)=1$
$f(1)=2$
$f(2)=3$
Since, the equation is non-homogenous, it will consists of Two Parts.
- Finding solution of Associated Homogenous Recurrence Relation $f(n)=f(n-<!--\; -\; -->1)+f(n-<!--\; -\; -->3)$
For this the characteristic equation will be $x^3-<!--\; -\; -->x^2-<!--\; -\; -->1=0$
- Finding Particular Solution
one can find answer by inputting $f(n)=f(n-<!--\; -\; -->1)+f(n-<!--\; -\; -->3)+1,f(0)=1,f(1)=2,f(2)=3$
The solution is quite complex. Thus, I was interested if we can deduce that $f(n)\ge <!--\; \ge \; -->x^n$ for some $x\mathrm{ϵ<!--\; ϵ\; -->}\mathbb{R}$.

Find k: ${27}^k\equiv <!--\; \equiv \; -->2\mathrm{mod}2021$
By using these: $2^{11}=2048\equiv <!--\; \equiv \; -->27\mathrm{mod}2021$
$2021=43\cdot <!--\; \cdot \; -->47$
I should find integer k
${27}^k\equiv <!--\; \equiv \; -->2\mathrm{mod}2021$
Well, I can only come up with something like this:
$a^{42}\equiv <!--\; \equiv \; -->1\mathrm{mod}43$
$b^{46}\equiv <!--\; \equiv \; -->1\mathrm{mod}47$
$a\ne <!--\; \ne \; -->0\mathrm{mod}43,b\ne <!--\; \ne \; -->0\mathrm{mod}47$

Need a clarification of the proof that the prime ideal space of a distributive bounded lattice is compact
Let L be a bounded distributive lattice, then the prime ideal space $⟨<!--\; ⟨\; -->{\mathcal{I}}_p(L);\mathrm{\tau <!--\; \tau \; -->}⟩<!--\; ⟩\; -->$ is compact. $\mathrm{\tau <!--\; \tau \; -->}$ is the topology whose basis is $\mathcal{B}=\{X_b\cap <!--\; \cap \; -->(X\setminus <!--\; \setminus \; -->X_c):b,c\in <!--\; \in \; -->L\}$, with $X_a=\{I\in <!--\; \in \; -->{\mathcal{I}}_{\mathcal{p}}(L):a\notin <!--\; \notin \; -->I\}$.
We want to prove that the subbasis $\mathcal{S}=\{X_b:b\in <!--\; \in \; -->L\}\cup <!--\; \cup \; -->\{X\setminus <!--\; \setminus \; -->X_c:c\in <!--\; \in \; -->L\}$ satisfies Alexander's Lemma. Let $\mathcal{U}:=\{X_b:b\in <!--\; \in \; -->A_0\}\cup <!--\; \cup \; -->\{X\setminus <!--\; \setminus \; -->X_c:c\in <!--\; \in \; -->A_1\}$, a open cover of ${\mathcal{I}}_{\mathcal{p}}(L)$.
Let J be the ideal generated by $A_0$ and G the filter generated by $A_1$. It is easy to prove that $J\cap <!--\; \cap \; -->G\ne <!--\; \ne \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}$. So $J\cap <!--\; \cap \; -->G\ne <!--\; \ne \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}$ and let $a\in <!--\; \in \; -->J\cap <!--\; \cap \; -->G$; if $A_0$ and $A_1$ are both non-empty, there exist $b_1,\dots <!--\; \dots \; -->,b_j\in <!--\; \in \; -->A_0$ and $c_1,\dots <!--\; \dots \; -->,c_k\in <!--\; \in \; -->A_1$ s.t. $c_1\wedge <!--\; \wedge \; -->\cdots <!--\; \cdots \; -->\wedge <!--\; \wedge \; -->c_k\le <!--\; \le \; -->a\le <!--\; \le \; -->b_1\vee <!--\; \vee \; -->\cdots <!--\; \cdots \; -->\vee <!--\; \vee \; -->b_j$, whence $X=X_1=X_{b_1}\cup <!--\; \cup \; -->\cdots <!--\; \cdots \; -->\cup <!--\; \cup \; -->X_{b_j}\cup <!--\; \cup \; -->(X\setminus <!--\; \setminus \; -->X_{c_1})\cup <!--\; \cup \; -->\cdots <!--\; \cdots \; -->\cup <!--\; \cup \; -->(X\setminus <!--\; \setminus \; -->X_{c_k})$
What escapes me is why we can write 1 in that way if $J\cap <!--\; \cap \; -->G\ne <!--\; \ne \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}$?

How to calculate summation of $\frac{n^2+n+1}{n^2+n}$ over 1 to 25?
Can we calculate the summation of numerator, and denominator and then divide them both? But that does not seems to be a plausible way. So, I reduced it to $S=1+\frac{1}{n^2+n}$. But now I do not how to compute this sum.

Find smallest positive $a\in <!--\; \in \; -->\mathbb{Z}$ such $10a\equiv <!--\; \equiv \; -->7\mathrm{mod}17$
I found that the answer was $a=16$, I believe.
Note that $10a\equiv <!--\; \equiv \; -->7\mathrm{mod}17⟺<!--\; ⟺\; -->17\mid <!--\; \mid \; -->10a-<!--\; -\; -->7⟺<!--\; ⟺\; -->10a-<!--\; -\; -->7=17x,x\in <!--\; \in \; -->\mathbb{Z}$
The way I went about it was running EEA for $10a+17(-<!--\; -\; -->x)=7,x\in <!--\; \in \; -->\mathbb{Z}.$.
I found it to be $1=10(-<!--\; -\; -->5)+17(3)⟹<!--\; ⟹\; -->7=10(-<!--\; -\; -->35)+17(21)$. But I wasn't sure where then I could find 16 for this. I ended using bruteforce to find $a=16$ starting from 1. (not efficient).
Any more efficient alternative to find this such a is appreciated!

Binary strings with exactly n ones but without 000 or 111
I am a newbie in binary strings and generating series. I have a problem of enumeration which can be transformed to the following: find the number of binary strings of exactly n ones which do not contain three consecutive ones or three consecutive zeros. The answer can be given as a coefficient of a generating series.

How to solve $x(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)(x-<!--\; -\; -->3)(x-<!--\; -\; -->4)(x-<!--\; -\; -->5)....(x-<!--\; -\; -->999)$

How do I prove $B=C$ when...
If $A\cap <!--\; \cap \; -->B=A\cap <!--\; \cap \; -->C$ and $A^{\mathrm{\complement <!--\; \complement \; -->}}\cap <!--\; \cap \; -->C$, then $B=C$.
I seem to be able to prove this without the use of $A^{\mathrm{\complement <!--\; \complement \; -->}}\cap <!--\; \cap \; -->C$ but it must be there for a reason. Where would I add this in. The basic gist behind what I did was say x was an element of A therefore x is in $A\cap <!--\; \cap \; -->B$ and $A\cap <!--\; \cap \; -->C$. So x is in B and x is in C. So since they share the same element they are equal. This may not be right but I think I'm somewhere in the ballpark. Any help would be greatly appreciated.

At least how many residents of the country have the same birthday?
Let's suppose that a country has 11.000.000 people. At least how many residents of the country have the same birthday?
Take into consideration that leap years exist.
What I did is:
Let's suppose we have 2021. I take $\frac{\mathrm{10.000.000}\cdot <!--\; \cdot \; -->2021}{\mathrm{11.000.000}}=1838$.

Number of functions of k-arity
I want to find the number of possible functions $g_i$ acting on k elements $(p_1,\dots <!--\; \dots \; -->,p_k)$ if the possible values of $p_j$s and $g_i$ are $(v_1,\dots <!--\; \dots \; -->,v_n)$.
My thoughts: $g_i:\{v_1,\dots <!--\; \dots \; -->,v_n{\}}^k\to <!--\; \to \; -->\{v_1,\dots <!--\; \dots \; -->,v_n\}$, $g_i:(p_1,\dots <!--\; \dots \; -->,p_k)\mapsto <!--\; \mapsto \; -->g_i(p_1,\dots <!--\; \dots \; -->,p_k)$. It's possible to imagine a tabular with each possible value of the input and so having in total $n^k$ rows. Now the values that $g_i$ can take are again between n option for each row, and so I get $n^{kn}$ possible function. Is that true?

Can't solve $(A\setminus <!--\; \setminus \; -->B)\setminus <!--\; \setminus \; -->C=(A\setminus <!--\; \setminus \; -->C)\setminus <!--\; \setminus \; -->(B\setminus <!--\; \setminus \; -->C)=A\setminus <!--\; \setminus \; -->(B\cup <!--\; \cup \; -->C)$ any help?
I have discrete math exam and i can't quite figure out one example.
$(A\setminus <!--\; \setminus \; -->B)\setminus <!--\; \setminus \; -->C=(A\setminus <!--\; \setminus \; -->C)\setminus <!--\; \setminus \; -->(B\setminus <!--\; \setminus \; -->C)=A\setminus <!--\; \setminus \; -->(B\cup <!--\; \cup \; -->C)$
i tried solving it like this and i got stuck: $(A\setminus <!--\; \setminus \; -->B)\setminus <!--\; \setminus \; -->C=(A\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{B})\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{C}$, what next ?

Prove that $n\in <!--\; \in \; -->\mathbb{N}$ and $n\le <!--\; \le \; -->3$, then $2^n<3n+2$.
Prove that $n\in <!--\; \in \; -->\mathbb{N}$ and $n\le <!--\; \le \; -->3$, then $2^n<3n+2$
Need to prove this sentence is true. In this problem, I believe I have to use the exhaustion demo technique. Because there is a specific number of cases, in this case between natural 0 and 3.I've tried to solve it, I came to a conclusion, and apparently it's correct. But I would like someone to look at the strategy used for the resolution and if I am correct about my conclusion.
My resolution method:
Create a table with possible values for n and compare the sentence
$\begin{array}{ccl}n& 2^n<3n+2& \text{result}\\ 0& 1<2& \text{true}\\ 1& 2<5& \text{true}\\ 2& 4<8& \text{true}\\ 3& 8<11& \text{true}\end{array}$
Thus, by exhaustion, it is possible to prove that the theorem is true. For any and all $n\ge <!--\; \ge \; -->0$ and $\le <!--\; \le \; -->3$, it will always be less than $2^n<3n+2$.

Let P(n) be an open statement about $n\in <!--\; \in \; -->\mathbb{N}$. In the weak induction, beside the base case, it is stated:
$\mathrm{\forall <!--\; \forall \; -->}k\in <!--\; \in \; -->\mathbb{N}(P(k)⟹<!--\; ⟹\; -->P(k+1))$ (i)
But, in this page right in the beginning, it states differently, according to my understanding (with $a=1$)
$((\mathrm{\exists <!--\; \exists \; -->}k\in <!--\; \in \; -->\mathbb{N})P(k))⟹<!--\; ⟹\; -->P(k+1)$ (ii)
Are the statements (i) and (ii) equivalent? If yes, why?

How many ways we can choose an increasing and a strictly increasing subsequence of length k from the first n natural numbers?
Let $S:=\{1\dots <!--\; \dots \; -->n\}.$. We want to choose
i) increasing subsequences of S length $k\le <!--\; \le \; -->n.$. and
ii) strictly increasing subsequences of S length $k\le <!--\; \le \; -->n.$.
How many ways can we choose them in i) and ii)? I'm a bit confused where even to start, but detailed answers will be appreciated!

How can I demonstrate a succession with a floor function by induction?
Show that for $n\ge <!--\; \ge \; -->1$ it is satisfied that.
$\lfloor <!--\; \lfloor \; -->\frac{1}{2}\rfloor <!--\; \rfloor \; -->+\lfloor <!--\; \lfloor \; -->\frac{2}{2}\rfloor <!--\; \rfloor \; -->+\lfloor <!--\; \lfloor \; -->\frac{3}{2}\rfloor <!--\; \rfloor \; -->+\cdots <!--\; \cdots \; -->+\lfloor <!--\; \lfloor \; -->\frac{n}{2}\rfloor <!--\; \rfloor \; -->=\lfloor <!--\; \lfloor \; -->\frac{n}{2}\rfloor <!--\; \rfloor \; -->\lfloor <!--\; \lfloor \; -->\frac{n+1}{2}\rfloor <!--\; \rfloor \; -->$
where ⌊x⌋ is the lower integer part of x (floor function evaluated at x).
any idea to start to make it.

Is it true that for every regular language $L\subseteq <!--\; \subseteq \; -->\{0,1{\}}^{\ast <!--\; \ast \; -->}$ the language $\{w^{|w|}|w\in <!--\; \in \; -->L\}$ is also regular?

How many relations are there on a set with n elements?
Discrete Mathematics and Its Applications 8th Edition by Kenneth Rosen states the following on Pg 600:
...a binary relation from A to B is a set R of ordered pairs, where the first element of each ordered pair comes from A and the second element comes from B.
So if A has 3 elements and B has 3 elements, the binary relation should have $3\cdot <!--\; \cdot \; -->3=9$ elements.
Now let us consider the case where we have a binary relation from A to itself. In this case also I expect the binary relation to have $3\cdot <!--\; \cdot \; -->3=9$ elements.
However, on Pg 602 the author states that:
there are $2^{3^2}=2^9=512$ relations on the set (a, b, c).
I am quite confused by this and would appreciate any clarification.