If x, y, z,.................... are (m + 1) distinct prime numbers, the number of factors of ${x^n}yz...............$ where $\left( {n \in N} \right)$ is
$
  (a){\text{ m(n + 1)}} \\
  {\text{(b) }}{{\text{2}}^n}m \\
  (c){\text{ (n + 1)}}{{\text{2}}^m} \\
  (d){\text{ n}}{\text{.}}{{\text{2}}^m} \\
$

Question

If x, y, z,.................... are (m + 1) distinct prime numbers, the number of factors of ${x^n}yz...............$ where $\left( {n \in N} \right)$ is
$
  (a){\text{ m(n + 1)}} \\
  {\text{(b) }}{{\text{2}}^n}m \\
  (c){\text{ (n + 1)}}{{\text{2}}^m} \\
  (d){\text{ n}}{\text{.}}{{\text{2}}^m} \\
$

Vedantu Content Team · Accepted Answer

Hint – In this problem first find the total number of factors of ${x^n}$, considering x as a prime number and n as natural number, then evaluate the factors of yz……………………. . . . . Remember that the series \[yz...............\] will go up to m as x, y, z,.................... are (m + 1) distinct prime numbers. The overall factors will be the multiplication of factors of ${x^n}$ and series of yz……………………. respectively.
Given distinct prime numbers are
x, y, z, .................... are (m + 1) distinct prime numbers.
As we know that prime numbers have only two factors (1 and itself).
So all these distinct prime numbers have 2 factors.
Now we have to calculate the number of factors of ${x^n}yz...............$ where $\left( {n \in N} \right)$.
So find out the number of factors of $x^n$.
As n belongs to natural number so n =0, 1, 2, 3, 4, .........., n
So the number of factors of $x^n$ are $ (1, x, x^2, x^3,....x^n) $.
So the number of factors of $x^n$ are (n + 1).
And factors of y, z, ........, m (as the number of distinct prime numbers remains (m) as we calculated the factors of xn).
So the factors of y, z, ......, m = 2, 2, 2, ........., 2
As there are m terms so the number of factors = $2^m$.
So the total number of factors of ${x^n}yz = \left( {n + 1} \right){2^m}$.
So this is the required answer.
Hence option (C) is correct.

Note – The key point here was the evaluation of factors of the series y, z, ........, m, since the definition of prime number itself suggests that a prime number is one which is divisible by 1 and itself, this means that a prime number has exactly 2 factors. This if 1 number has 2 factors and in total we have m numbers in the series then the total number of factors of that series will be ${2^m}$.

If x, y, z,.................... are (m + 1) distinct prime numbers, the number of factors of ${x^n}yz...............$ where $\left( {n \in N} \right)$ is$ (a){\text{ m(n + 1)}} \\ {\text{(b) }}{{\text{2}}^n}m \\ (c){\text{ (n + 1)}}{{\text{2}}^m} \\ (d){\text{ n}}{\text{.}}{{\text{2}}^m} \\ $

If x, y, z,.................... are (m + 1) distinct prime numbers, the number of factors of ${x^n}yz...............$ where $\left( {n \in N} \right)$ is
$
(a){\text{ m(n + 1)}} \\
{\text{(b) }}{{\text{2}}^n}m \\
(c){\text{ (n + 1)}}{{\text{2}}^m} \\
(d){\text{ n}}{\text{.}}{{\text{2}}^m} \\
$