Question

The number of positive factors of 9600 including 1 and 9600 are
A.60
B.58
C.48
D.46

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 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 9600.
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 9600 by 2 as it’s the first prime number.
$ \Rightarrow \frac{{9600}}{2} = 4800$
We get 4800 which is a even number so we can divide it further by 2
$ \Rightarrow \frac{{4800}}{2} = 2400$
Now we get 2400 which even so let's proceed by dividing it by 2
$ \Rightarrow \frac{{2400}}{2} = 1200$
We get 1200 which is a even number so we can divide it further by 2
$ \Rightarrow \frac{{1200}}{2} = 600$
We get 600 which is a even number so we can divide it further by 2

$ \Rightarrow \frac{{600}}{2} = 300$
We get 300 which is a even number so we can divide it further by 2

$ \Rightarrow \frac{{300}}{2} = 150$
We get 150 which is a even number so we can divide it further by 2

$ \Rightarrow \frac{{150}}{2} = 75$
Now we have 75 which is not divisible by 2
So $9600 = 2*2*2*2*2*2*2*75 = {2^7}*75$
Step 2 :
Here 75 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 75 by 3
$ \Rightarrow \frac{{75}}{3} = 25$
We get 25 which is not divisible by 3 .
So now $9600 = 2*2*2*2*2*2*2*3*25 = {2^7}*3*25$
Step 3:
Now 25 is not a prime number , so let's divide it by the next prime number 5
$ \Rightarrow \frac{{25}}{5} = 5$
Now we get 5 and we already know 5 is a prime number
Hence we cannot divide it further.
Therefore , $9600 = 2*2*2*2*2*2*2*3*5*5 = {2^7}*{3^1}*{5^2}$
Step 4:
Now we have expressed 9600 as the product of primes.
Therefore when a number N is expressed as $N = {p^a}*{q^b}*{r^c}$ , where p,q,r, are primes.
The number of factors of N is given by $\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)$
Here p = 2 , q = 3 , r = 5 and a = 7 ,b = 1 ,c = 2
Therefore the number of factors of 9600 is ,
$
   \Rightarrow \left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right) \\
   \Rightarrow \left( {7 + 1} \right)\left( {1 + 1} \right)\left( {2 + 1} \right) \\
   \Rightarrow 8*2*3 \\
   \Rightarrow 48 \\
$
The correct option is C

Note: Points to remember: 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.