MCQ
Let $N_1=2^{55}+1$ and $N_2=165$.Then
- A$N_1$ and $N_2$ are coprime
- Bthe $HCF$ (Highest Common Factor) of $N_1$ and $N_2$ is $55$
- Cthe $HCF$ of $N_1$ and $N_2$ is $11$
- ✓the $HCF$ of $N_1$ and $N_2$ is $33$
It is given that, $N_2=165$ $=3 \times 5 \times 11$ and $N_1=2^{55}+1$
As we know that, if $n$ is odd integer then $x^n+y^n$ is divisible by $x+y$.
So, $N_1=2^{55}+1^{55}$ is divisible by $2+1=3$ and $N_1=2^{55}+1^{55}$
$=\left(2^5\right)^{11}+\left(1^5\right)^{11}=(32)^{11}+(1)^{11}$
is divisible by $32+1=33$
$\therefore$ the HCF of $N_1$ and $N_2$ is $33$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ The equation of circle $\mathrm{C}$ is
$(A)$ $(x-2 \sqrt{3})^2+(y-1)^2=1$
$(B)$ $(x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1$
$(C)$ $(x-\sqrt{3})^2+(y+1)^2=1$
$(D)$ $(x-\sqrt{3})^2+(y-1)^2=1$
$2.$ Points $E$ and $F$ are given by
$(A)$ $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$
$(B)$ $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right),(\sqrt{3}, 0)$
$(C)$ $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$(D)$ $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$3.$ Equation of the sides $Q R, R P$ are
$(A)$ $y=\frac{2}{\sqrt{3}} x+1, y=-\frac{2}{\sqrt{3}} x-1$
$(B)$ $y=\frac{1}{\sqrt{3}} x, y=0$
$(C)$ $y=\frac{\sqrt{3}}{2} x+1, y=-\frac{\sqrt{3}}{2} x-1$
$(D)$ $y=\sqrt{3} x, y=0$
Give the answer question $1,2$ and $3.$