How can i quickly find the factors of a particular number?
Find the number of different factors of 1800 and 3003?
This being the question , for 3003 i first found out its prime factors and then i could find the factors by multiplying each other.......but this was a tedious method and when i started doing it for 1800 ,i ended up dumbstruck!
So, plz help me by giving a quick way of doing this!
If $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, where the $p_i$ are distinct primes, then the number of (positive) factors of $n$ is $$(a_1+1)(a_2+1)\cdots (a_k+1).$$
For example, if $n=1800$, then since $1800=2^3\cdot 3^2\cdot 5^2$, the number of positive factors of $n$ is $(3+1)(2+1)(2+1)$.
Remarks: $1.$ The above formula is easy to use if we know the prime power factorization of $n$. However, if $n$ is large, then finding that factorization can be computationally difficult.
$2.$ Here is an outline of a proof. We are trying to make a factor $d$ of $n$. We first look at $p_1$ and decide how many $p_1$'s $d$ will get. The choices are $0$ $p_1$'s, or $1$, or $2$, and so on up to $a_k$, a total of $a_1+1$ choices. For every such choice, there are $a_2+1$ ways to decide how many $p_2$'s the factor $d$ will get. And so on.