Question

If x, y, z,.........are (m + 1) distinct prime numbers, the numbers of factors of $x^{n}yz......\left(nϵN\right)$ is

• A. m (n + 1)
• B. $2^{n}m$
• C. $(n+1)2^m$
• D. $n.2^m$

Answer 1

The correct option is C $(n+1)2^m$

Given $x,y,z......$ are $(m+1)$ distinct prime nos

Now $x^{n}yz............\left(nϵN\right)$

$x^n$ has $\frac{(n+1)}{2}$ factors ($\because$ factors of $x^n=x,x^2,x^3....x^n$& $1$)

& $y,z,......$ has $2$ factors (one and itself)

$\therefore$ No. of factors of $x^{n}yz.....$ are

$(n+1)\mathrm{2.2........}m+1$

$\therefore (n+1)2^m$