I thought that the number of divisors of a number was the product of the indices in its factorisation, plus $2$ (for 1 and the number itself). For instance, $2000=2^{4} \cdot 5^{3}$, so it would have $4 \cdot 3 + 2 = 14$ divisors.
Apparently, however, $2000$ actually has $20$ divisors. What am I doing wrong?
On
You take $2$ to any power between $0$ and $4$, and $3$ with any power between $0$ and $3$, hence $5\times4$ combinations.
You forgot the options such that one exponent is zero, $2,4,8,16$ and $5,25,125$, and you counted $2^4\cdot5^3=2000$ twice. Hence, $14-1+4+3=20$.
On
A divisor of $2^4\,5^3$ has the prime decomposition $2^i\,5^j$, where $\;0\le i\le 4$ ($5$ possibilities) and $0\le j\le3$ ($4$ possibilities). Combining these possibilities, this make $5\cdot 4$ divisors in all.
More generally, given the prime decomposition of a number $n$, the number of its positive divisors is the product of the exponents plus $1$ of its prime factors.
On
Brute force: \begin{array}{c|cccc}1&2&4&8&16\\\hline5&10&20&40&80\\25&50&100&200&400\\125&250&500&1000&2000\end{array} In general, if $n=\prod_i p_i^{a_i}$ then the number of divisors is $\tau(n)=\prod_i(a_i+1)$.
On
Lets look at any natural number and it's positive divisors.
Every number $n$ greater than $1$ can be represented uniquely in a form of $$n = (p_1)^{i_1}\cdot(p_2)^{i_2}\cdot\ldots\cdot(p_m)^{i_m}=\prod_{1\leq l \leq m} (p_l)^{i_l}$$ where $p_l$ is a strictly growing sequence of prime numbers.
Now any number that is in a form $$d = (p_1)^{j_1}\cdot(p_2)^{j_2}\cdot\ldots\cdot(p_m)^{j_m}=\prod_{1\leq l \leq m}(p_l)^{j_l}, \text{where } \forall_{1\leq l\leq m} 0\leq j_l \leq i_l$$ is a divisor of $n$ and all divisors of $n$ are exactly in that form. The proof is obvious so let me skip it.
In other words we have $l$ sets of natural numbers from $0$ to $i_l$ so each of those sets has exactly $i_l+1$ elements. To create a divisor of $n$ we need to pick exactly one element from each of the sets. We can do that in $$\sigma=\prod_{1\leq l\leq m}(i_l+1) = (i_1+1)\cdot(i_2+1)\cdot\ldots\cdot(i_m+1)$$ So $\sigma$ is your total number of unique positive divisors of $n$.
In your case $$2000=2^4\cdot5^3$$ hence $$m=2$$ $$p_1=2$$ $$p_2=5$$ $$i_1=4$$ $$i_2=3$$ $$\sigma = (4+1)\cdot(3+1)=5\cdot 4 = 20$$
What am I doing wrong
Well, the answer is actually here
I thought that the number of divisors of a number was the product of the indices in its factorisation, plus 2 (for 1 and the number itself)
With this you ignore the possibility of entirely dropping $l\text{-th}$ prime number when creating your divisors (i.e. having $j_l=0$ for any $l$).
The factors are of the form of $2^x\cdot 5^y$ where $x$ takes value from $0$ to $4$ and $y$ takes values from $0$ to $3$.
Hence there is a total of $(4+1)(3+1)=20$ factors.
Your method miss out number such as $5^y, y>0; 4^x, x>0$ and double counted $2000$.