Question

The number of positive factors of 2520 excluding unity is
A.48
B.45
C.46
D.47

Answer 1

Hint: To find the number of factors of a given number, express the number as a product of powers of prime numbers. Now, increment the power of each of the prime numbers by 1 and multiply the result. That is we need to write the number N in the form of $N = {p^a}*{q^b}*{r^c}$ where p , q , r are prime numbers then the number of factors excluding 1 is given by the formula $\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)$

Complete step-by-step answer:
Step 1:
The given number is 2520.
Firstly, we need to express this into the product of primes.
Let's do prime factorization to express 2520 as the product of primes
So let's divide 2520 by 2 as it’s the first prime number.
$ \Rightarrow \frac{{2520}}{2} = 1260$
We get 1260 which is a even number so we can divide it further by 2
$ \Rightarrow \frac{{1260}}{2} = 630$
Once we get 630 which even so let's proceed by dividing it by 2
$ \Rightarrow \frac{{630}}{2} = 315$
Now we have 315 which is not divisible by 2
So $2520 = 2*2*2*315 = {2^3}*315$
Step 2 :
Here 315 is not a prime number so we can continue factoring it with the next prime number.
The next prime number is 3
So let's divide 315 by 3
$ \Rightarrow \frac{{315}}{3} = 105$
We get 105 which is divisible by 3 so let's divide it by 3 again
$ \Rightarrow \frac{{105}}{3} = 35$
We get 35 which is not divisible by 3 .
So now $2520 = 2*2*2*3*3*35 = {2^3}*{3^2}*35$
Step 3:
Now 35 is not a prime number , so let's divide it by the next prime number 5
$ \Rightarrow \frac{{35}}{5} = 7$
Now we get 7 and 7 is a prime number
Hence we cannot divide it further.
Therefore , $2520 = 2*2*2*3*3*5*7 = {2^3}*{3^2}*5*7$
Step 4:
Now we have expressed 2520 as the product of primes.
Therefore when a number N is expressed as $N = {p^a}*{q^b}*{r^c}*{s^d}$ , where p,q,r,s are primes.
The number of factors of N excluding 1 is given by $\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)\left( {d + 1} \right)$
Here p = 2 , q = 3 , r = 5, s = 7 and a = 3 ,b = 2 ,c = 1 and d = 1
Therefore the number of factors of 2520 excluding 1 is ,
$
   \Rightarrow \left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)\left( {d + 1} \right) \\
   \Rightarrow \left( {3 + 1} \right)\left( {2 + 1} \right)\left( {1 + 1} \right)\left( {1 + 1} \right) \\
   \Rightarrow 4*3*2*2 \\
   \Rightarrow 48 \\
$

The correct option is A

Note: 1.Every number is a factor of itself.
2.Every factor of a given number is either less than or equal to the given number.
3.1 is a factor of every given number.
4.Every factor of a given number is an exact divisor of that given number.
5.Number of factors of a given number are finite