The task is to find out how many positive integers divide at least one of the given numbers: $10^{60}$, $20^{50}$, $30^{40}$. This is easily calculated using the inclusion-exclusion principle.
First, we have to figure out the form of divisors for each of these numbers.
$10^{60}=(2\cdot5)^{60}=2^{60}\cdot5^{60}$, which means divisors of this number are of form $2^{i}\cdot5^{j}$ where $0\le{i},j\le{60}$
$20^{50}=(2\cdot2\cdot5)^{50}=2^{2\cdot50}\cdot5^{50}$, divisors: $2^{2\cdot{i}}\cdot5^{j}$ where $0\le{i},j\le{50}$
$30^{40}=(2\cdot3\cdot5)^{40}=2^{40}\cdot3^{40}\cdot5^{40}$, divisors: $2^{i}\cdot3^{j}\cdot5^{k}$ where $0\le{i},{j},{k}\le{40}$
Number of divisiors of form $2^{i}\cdot5^{j}$ where $0\le{i},j\le{60}$ equals $61\cdot61$, because we have 61 combinations for both $i$ and $j$. Following this logic, number of divisiors of form $2^{2\cdot{i}}\cdot5^{j}$ where $0\le{i},j\le{50}$ should be $51\cdot51$, because we have 51 choices for $i$ and $j$. But, the solution I read said the number of these divisors is $101\cdot51$. I feel like this has something to do with $2^{2\cdot{i}}$, although this equals to $4^{i}$ and shouldn't change anything. Could the solution be wrong?
Thank you in advance.
Note that $2^{2*50}=2^{100}$. Thus, there are $101$ such numbers of the form $2^j$ such that $0 \leq j \leq 100$. You reasoned about this the right way but got a little too caught up! The first and last problems behave similarly since their exponents were such that $i=j=k$.
Just to complete the problem: Using the reasoning you mentioned, we have that there are $101*51$ ways to make numbers of the form $2^j5^k$ such that $0\leq j \leq 100$ and $0 \leq k \leq 50$.