Question

If the $\left( {n + 1} \right)$ numbers $a,b,c,d, \cdots $ be all different and each of them a prime number, then find the number of different factors (other than 1) of ${a^m} \cdot b \cdot c \cdot d \cdots $.
A. $\left( {m + 1} \right){2^n}$
B. $\left( {m + 1} \right){2^n} - 1$
C. $\left( {m + 1} \right){2^n} + 1$
D. None of these

Answer 1

Hint: First we find the prime factors of ${a^m}$ by using the theorem that the prime factor of ${a^p}$ is $\left( {p + 1} \right)$. We know that a prime factor has 2 factors that are itself and 1. By using this we will find the prime factors of $a, b,c,d, \cdots $. Since all numbers are prime numbers, they have no common factor except 1. Multiply the number prime factor of ${n + 1}$ numbers and subtract 1 to get the desired result.

Complete step by step solution:
Given $\left( {n + 1} \right)$ numbers are $a,b,c,d, \cdots $ and all numbers are distinct.
The number of prime factors of ${a^m}$ is $\left( {m + 1} \right)$.
The number of prime factors of each prime number 2.
Since each prime number is distinct, there is no common factor of prime numbers.
Now we will calculate the number of prime factors of ${a^m} \cdot b \cdot c \cdot d \cdots $
The prime factors of ${a^m} \cdot b \cdot c \cdot d \cdots $ is $\left( {m + 1} \right) \cdot 2 \cdot 2 \cdot 2 \cdots n\,{\rm{times}}$
                                                                    $ = \left( {m + 1} \right){2^n}$
Since 1 is a factor of ${a^m} \cdot b \cdot c \cdot d \cdots $. So, we subtract 1 from the total number of factors.
The number of different factors (other than 1) of ${a^m} \cdot b \cdot c \cdot d \cdots $ is $ = \left( {m + 1} \right){2^n} - 1$

Option ‘B’ is correct

Note: Be careful while calculating the number of factors make sure you subtract 1 from the total number of factors i.e., here $\left( {m + 1} \right){2^n}$. Do not skip the point that we have to calculate the number of prime factors except 1.