Positive integers are divisors

The given numbers can be factored as $10^{40} = 2^{40}5^{40}$ and $20^{30} = (2^{2}5)^{30} = 2^{60}5^{30}.$

The positive integers that we are then looking for are those that either

• (i) divide the GCD of the given values, i.e., that divide $2^{40}5^{30},$

• (ii) divide $2^{40}5^{40}$ but not $2^{60}5^{30},$ or

• (iii) divide $2^{60}5^{30}$ but not $2^{40}5^{40}.$

Option (i) provides us with $(40 + 1)(30 + 1) = 1271$ divisors.

For option (ii) we have those numbers of the form $2^{m}5^{n}$ where $0 \le m \le 40$ and $31 \le n \le 40.$ This provides us with $(40 + 1)(40 - 31 + 1) = 410$ more divisors.

For option (iii) we have those numbers of the form $2^{m}5^{n}$ where $41 \le m \le 60$ and $0 \le n \le 30.$ This provides us with $(60 - 41 + 1)(30 + 1) = 620$ more divisors.

Thus the total number of divisors being sought after is $1271 + 410 + 620 = \boxed{2301}.$

Note that this problem is simply asking us to find the $(\text{number of divisors of } 10^{40}) + (\text{number of divisors of } 20^{30}) - (\text{number of divisors of } \text{gcd}(10^{40},20^{30}))$ since the divisors of $\text{gcd}(10^{40},20^{30})$ appear once in each of the divisors of $10^{40}$ and the divisors of $20^{30}$ .

Consider that $10^{40} = 2^{40}5^{40}$ , $20^{30} = 4^{30}5^{30} = 2^{60}5^{30}$ , and $\text{gcd}(10^{40},20^{30}) = 2^{40}5^{30}$

Therefore the answer we are looking for is $(40+1)(40+1)+(60+1)(30+1)-(40+1)(30+1) = 41^{2}+(61)(31)-(41)(31) = 1681+1891-1271 = \boxed{ 2301 }$