Divisors of 2016

Observe That :

Prime Factorization of 2016 is $\Rightarrow$ $2^{5} \times 3^{2} \times 7^{1}$

By Divisors of an integer Formula , we get

$(5+1)(2+1)(1+1)$ divisors

$6 \times 3 \times 2$ divisors

so , $2016$ has $\boxed{36}$ positive Divisors

2016= $2^{5}*3^{2}*7$

Formula for number of positive divisors for a number N= $a^{p}*b^{q}*c^{r}$ is (p+1)(q+1)(r+1).

So the answer is (5+1)(2+1)(1+1)= $\boxed{36}$

$2016 = 2^5 * 3^2 * 7^1$

Therefore, we can get a divisor of 2016 taking $2^x * 3 ^y * 7^z$ where $x$ is any value from $0$ to $5$ , $y$ is any value from $0$ to $2$ and $z$ is any value from $0$ to $1$ (integer values).

So, our answer is $(5+1)(2+1)(1+1)$ Note that it is $5 + 1$ and not $5$ because we are including $0$ .

$=\boxed{36}$

Using prime factorization, $2016=2^5\times3^2\times7$ .

Therefore, $2016$ has $(5+1)(2+1)(1+1) = 36$ positive divisors.

2016 = 2^{5} \times 3^{2} \times 7^{1} You form a divisor by picking a subset of the prime factors to multiply together. There are 6 choices for how many 2's to include (0, 1, 2, 3, 4, or 5), 3 choices for how many 3's to include (0, 1, or 2), and 2 choices for how many 7's to include (0 or 1). Thus the number of divisors is 6 \times 3 \times 2 = \boxed{36}

there is a formula for this ..... n=p^a * q^b * r^c....... ( ^ is the symbol for power...ex. 2 2 2 2 = 2^4) d(n)=(a+1) * (b+1) * (c+1)........ let a no. be 2016=2 2 2 2 2 3 3 7 = 2^5 * 3^2 * 7^1 (according to equation n=p^a q^b r^c......a=5,b=2,c=1) d(2016)=(5+1)(2+1)(1+1) =6 3 2 =36

2016 can be factorized as (2^5) * (3^2) * (7)

Formula for finding total no. of divisors of (2^m) * (3^n) * (5^p) * (7^q) = (m + 1)(n + 1)(p + 1)(q + 1)

Therefore,

Total no. of divisors = 6 * 3 * 2 = 36 (Ans.)

2016 can be converted into product of three prime numbers namely 2, 3 and 7 with powers 5 , 2 and 1 respectively. So , Total no.s of factors=(5+1) (2+1) (1+1) =36.

The number of divisors a number has depends on what it's prime factors are so first we find the prime factors.

$2016$ is equal to $2^5 \cdot 3^2 \cdot 7^1$ in it's prime factorised form.

This doesn't give us all the divisors though but since we have it in it's factorised form the divisors become the product of varying amounts of the factors, for example if we ignore the $3^2$ and the $7^1$ you'll find that $32$ is a factor of $2016$ Since the powers can also be altered the amount of options available for each number become one more than the power (since it can also include zero) so the number of divisors of $2016$ is

$(5+1)(2+1)(1+1) = 36$

The general formula for this is

$n_{1}^{p_1} \cdot n_{2}^{p_2} \cdot n_{3}^{p_3}\cdot\ldots = n$

$(p_1 + 1)\cdot(p_2 + 1)\cdot(p_3 + 1)\cdot\ldots = d$

Where $n =$ Original number , $n_x =$ Prime factor of $n$ , $p_x =$ Power applied to $n_x$ and $d =$ Number of divisors

did it with this code

for(i=1;i<=n;i++) { if(n%i==0) { count++; } cout<<count;

2016=2^5 3^2 7 so 6 3 2=36

Brother Google.

First, find the prime factorization of 2016, which is 2 to the 5th power, 3 to the 2nd power, and 7. Then add 1 to each of the powers, and multiply them. 6 times 3 times 2 is 36.

a. First find the prime factorization: 2016 = 2^5 * 3^2 * 7^1 b. Adding 1 to each exponent we get: 5+1, 2+1 and 1+1 or 6,3,2 c. Multiplying these numbers together we get 36 d. The answer is 36.

The Prime Factorization of 2016 is: $$2^5\times 3^2 \times 7$$ The number of divisors of 2016: $$6\times 3 \times 2 = 36$$

2016=2^{5}.3^{2}.7 therefore number of divisors=(5+1)(2+1)(1+1)=6.3.2=36