Factor Counting

$\large 10^{10} = 2^{10} \cdot 5^{10}$

Now we know that if $N$ is a number such that it can be factorized in the form $p_1^{k_1} \cdot p_2^{k_2} \dotsm p_m^{k_m}$ where $p_i, \ 1 \leq i \leq m$ are distinct prime numbers and $k_i \in \mathbb Z, \ 1 \leq i \leq m$ , then the number of factors of $N$ is:

$(k_1+1)(k_2+1) \dotsm (k_m+1)$

Thus, applying the above result in $2^{10} \cdot 5^{10}$ , we get the number of factors of $10^{10}$ as $(10+1)(10+1) = 11 \cdot 11 = \boxed{121}$

The divisors are of the form $2^n5^m$ with $0\leq{n},$ $m\leq{10}$ ; there are $11^2=\boxed{121}$ of them.

If number is represented as a^p×b^q×c^r × ... where a, b, c, ... are prime factors of number then

total number of divisors = (p+1)(q+1)(r+1)...

10^10 = 2^10 × 5^10 total divisors = (10+1)(10+1) = 121