Maths

1012 $\le$  Number in the question $\le$  23421

Also, the number has to use digits {2, 3, 4, 5, 6} without repetition and the number has to be divisible by $\frac{55}{11*5}$  

As the number has to be divisible by both 5 and 11,

5 → once place

Let us make 4-digit such numbers first:

{2, 3, 4, 6} (digits are not be repeated)

A number is divisible by 11 it difference of sum of its digits at even places and sum of digits at odd place is 0 or multiple of 11.

$A=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$

$A^2=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=I+B,B=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$A^4=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]B^2\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=0$

$A^{2n}={\left(I+B\right)}^n$

I + nB + 0 + 0 +……….

$A^{2n}=I+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 6n\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6n\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$X\text{'}{A}^{k}x=\left[33\right]$

$=\left[111\right]A^k\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$

$\left[111\right]A^{2n}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left(k=2n\right)$

= $\left[1+1+6n+1\right]=\left[6n+3\right]=\left[33\right]n=5$

k = 10

x² – x – 4 = 0

$P_n={\alpha }^n-{\beta }^n$

$?=\frac{P_{15}{P}_{16}-P_{14}{P}_{16}-P_{15}^2+P_{14}{P}_{15}}{P_{13}{P}_{14}}$

$=\frac{\left(P_{15}-P_{14}\right)\left(P_{16}-P_{15}\right)}{P_{13}{P}_{14}}$

$=\frac{4P_{13}\cdot 4P_{14}}{P_{13}\cdot P_{14}}=16$

P_n =  αⁿ - βⁿ

$={\alpha }^{n-1}\cdot \alpha -{\beta }^{n-1}\cdot \beta$

$={\alpha }^{n-1}\left({\alpha }^2-4\right)-{\beta }^{n-1}\left(\beta -4\right)$

$P_n={\alpha }^{n+1}-{\beta }^{n+1}-4\left({\alpha }^{n-1}-{\beta }^{n-1}\right)$

$P_n=P_{n+1}-4P_{n-1}\Rightarrow P_{n+1}-P_n=4P_{n-1}$

The Interquartile range is also a measure of statistical dispersion that indicates the range within which middle 50% of dataset remains. It is the difference between the third and first quartile of the given dataset. Also known as IQR, it represents the length of the box which illustrates the spread of middle 50% data. All those data points that are either below Q1 - 1.5 x IQR or above Q3 + 1.5 x IQR are considered as outliers.

Q3 denotes the third Quartile which is the statistical measure to represent value below which 75% of the data falls in dataset when arranged in ascending order. It is the median of all the observations of the upper half of the data set i.e. 25% of the data. It is commonly used in conjunction with Q1 and median to provide the summary of distribution of dataset.Before calculating the result, it is important to arrange the data in increasing order.

Q1 denotes the first Quartile which is a statistical term to represent value below which 25% of data in dataset falls when arranged in ascending order. Q1 divides dataset into four equal parts which can easily analyze spread and data distribution. It is the median of all the observations of the lower half of the data set. However, it is important to arrange the data in increasing order for accurate results. 

$\frac{3x^2-9x+17}{x^2+3x+10}=\frac{5x^2-7x+19}{3x^2+5x+12}$

$1+\frac{2x^2-12x+7}{x^2+3x+10}=1+\frac{2x^2-12x+7}{3x^2+5x+12}$

$\left(2x^2-12x+7\right)\left(\frac{1}{x^2+3x+10}-\frac{1}{3x^2+5x+12}\right)=0$

$\underset{x\to 10}{lim}{\left(\frac{{\left(x+2cosx\right)}^3+2{\left(x+2cosx\right)}^2+3sin\left(x+2cosx\right)}{{\left(x+2\right)}^3\pm 2{\left(x+2\right)}^2+3sin\left(x+2\right)}\right)}^x$

$e^{\frac{100}{16+3sin2}\left[\frac{12-3\left(4\right)+8*1-8+3cos2-3cos2}{1}\right]}$ , using L.H.L Rule, e0 = 1

$2x^2-12x+7=0or3x^2+5x+12=x^2+3x+10$

$x^2+x+1=0$

Sum of roots = 6, no solution.