For a given $n$ , calculate $\sum_{i=1}^{n}{\lfloor \frac{n}{i} \rfloor}$ The main idea of this problem is there are maximum $2\sqrt{n}$ different values of $\lfloor \frac{n}{i} \rfloor$. [This is also known as Harmonic lemma] [But why?] click to hide Case 1: $x\le \sqrt{n}$ Hence, if we divide $n$ with $1\le i\le x$, number of values of $\lfloor \frac{n}{i} \rfloor$ will be at most $x$ (because, if all values were different, even then there will be $x$ values).

$Ax+By+Cz=P$ , $0\le x,y,z$ and $200 \le \frac{C}{gcd(A,B,C)}$. For given $A,B,C,P$ find number of triplet $(x,y,z)$. Solution: \begin{align} 200 &\le \frac{C}{gcd(A,B,C)} \newline C &\ge 200\times gcd(A,B,C) \newline \therefore C &\ge 200 \end{align} Hence, we will rewrite the eqation as $Ax+By+Cz=P \Longrightarrow Ax+By=P-Cz=P'$ We will iterate over all possible values of $P'$(there will be maximum $\frac{10^{8}}{200}$), and for each value of $P'$ we will find number of pair $(x,y)$ that satisfy $Ax+By=P'$ , and sum them up.

For given $n$ and $m$ print the value LLL...TTT where LLL is the leading $3$ digits of $n^{m}$ and TTT is trailing $3$ digits of $n^{m}$. Inside Math: For TTT we just use modular arithmetic and find $n^{m} \text{ (mod 1000)}$ Now, assume that , number of digit of $n^{m}$ is $x$. Hence LLL must be equals to $\lfloor \frac{n^{m}}{10^{x-3}}\rfloor$ How to find x ? click to hide \begin{align} 10^{x-1} \le n^{m} &< 10^{x} \text{ [Number of digit in }10^{x-1}\text{ is equals to }x\text{ ]}\newline log_{10}{10^{x-1}} \le log_{10}{n^{m}} &< log_{10}{10^{x}} \newline x-1 \le m \times log_{10}{n} &< x \newline x &\le m \times log_{10}{n} + 1 \newline \therefore\text{maximum possible integer value of }&x\text{ is }\lfloor m \times log_{10}{n} + 1 \rfloor\text{ or }\lfloor m \times log_{10}{n} \rfloor +1 \end{align} But how to calculate $\lfloor \frac{n^{m}}{10^{x-3}}\rfloor$ ?

How many triplet $(a,b,c)$ are there such that $0\le a,b,c \le L$ and $\sum_{x=0}^{\infty}{(ax^{2}+bx+c)\times (\frac{1}{10})^{x+1}} = \frac{p}{q}$ , for given $p,q,L$. Inside Math: Suppose, $A = \sum_{x=0}^{\infty}{ax^2\times (\frac{1}{10})^{x+1}}$ $B = \sum_{x=0}^{\infty}{bx\times (\frac{1}{10})^{x+1}}$ $C = \sum_{x=0}^{\infty}{c\times (\frac{1}{10})^{x+1}}$ Hence, $\sum_{x=0}^{\infty}{(ax^{2}+bx+c)\times (\frac{1}{10})^{x+1}} = A+B+C$ Now, we know(Actually we don’t know this, we have to prove this. We’ll try to prove this in some other section) that: $\sum_{x=0}^{\infty}{x^{2}r^{x+1}} = \frac{r^{2}(r+1)}{(r-1)^{3}} \tag{1}$ $\sum_{x=0}^{\infty}{xr^{x+1}} = \frac{r^{2}}{(r-1)^{2}} \tag{2}$