Leela and Madan pooled their music $CD's$ and sold them. They got as many rupees for each $CD$ as the total number of $CD's$ they sold. They share the money as follows: Leela first takes $10$ rupees, then Madan takes $10$ rupees and they continue taking $10$ rupees alternately till Madan is left out with less than $10$ rupees to take. Find the amount that is left out for Madan at the end, with justification.
(d)
Let the total number of CD's sold by the Leela and Madan together $=x$ Total money obtained by them
$=(x \times x)=x^2$
They divided $x^2$ in such that, $x^2=10$ (an odd number) $+$ a number less than $10$
$\Rightarrow \quad x=10 q+r \quad[\because 0 \leq r < 10]$
$\Rightarrow \quad x^2=(10 q+r)^2$
$\Rightarrow \quad x^2=100 q^2+20 q r+r^2$
$r^2=10$ (an odd number) $+$ a number less
than $10$
$r=16$ or $36$
$r^2=10+6$ or $3(10)+6$
Hence, the amount left for Madan at the end is $6$ rupees.
If $|x - 2| + |x - 3| = 7$, then $x =$
Let $r$ be a real number and $n \in N$ be such that the polynomial $2 x^2+2 x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be
The number of solutions of $\sin ^2 \mathrm{x}+\left(2+2 \mathrm{x}-\mathrm{x}^2\right) \sin \mathrm{x}-3(\mathrm{x}-1)^2=0$, where $-\pi \leq \mathrm{x} \leq \pi$, is....................
The sum of all integral values of $\mathrm{k}(\mathrm{k} \neq 0$ ) for which the equation $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$ in $x$ has no real roots, is ..... .
The number of integers $a$ in the interval $[1,2014]$ for which the system of equations $x+y=a$, $\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$ has finitely many solutions is