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: $\sum_{x=0}^{\infty}{x^{2}r^{x+1}} = \frac{r^{2}(r+1)}{(r-1)^{3}} \tag{1}$ $\sum_{x=0}^{\infty}{xr^{x+1}} = \frac{r^{2}}{(r-1)^{2}} \tag{2}$

$\pi(n) = \text{ number of prime number smaller than or equal to n}$ . $rub(n) = \text{ number of palindromic number smaller than or equal to n}$ . For a given $p$ and $q$ find maximum such $n$ so that, $\pi(n)\le \frac{p}{q}\times rub(n)$ Inside Math: $\pi(n) \approx \frac{n}{ln(n)}$ [Prime number approximation] $rub(n) \approx 2\sqrt{n}$ maximum value of $\frac{p}{q} = 42$ Hence , \begin{align} \pi(n) &\le \frac{p}{q}\times rub(n) \newline \Longrightarrow \frac{n}{ln(n)} &\le 42 \times 2\sqrt{n} \newline \Longrightarrow \frac{\sqrt{n}}{ln(n)} &\le 84 \newline \Longrightarrow n_{max} &\approx \boxed{1415344} \newline \end{align}

Statement: In this problem you are asked to calculate : $\sum_{i=L}^{R}{\sum_{j=0}^{i}{[\binom{i}{j} \text{ (mod 2)} \equiv 0]}}$ , for given $L$ and $R$. Inside Math: From the lucas' theorem we can state that, $\sum_{j=0}^{i}{[\binom{i}{j} \text{ (mod 2)}\equiv 1]} = 2^{f(i)}$ . [$f(i) = \text{ number of one in binary representation of } i$] Suppose , $S(x) = \sum_{i=0}^{x}{\sum_{j=0}^{i}{[\binom{i}{j} \text{ (mod 2)} \equiv 0]}}$ . Hence , our answer will be $S(R)-S(L-1)$

Basic Dp problem

insert minimum number of character in a given string after which the resulting string contain a subsequence abcdefgh...xyz