$$20^{20} = 2^{20} \times 2^{20} \times 5^{20} = 2^{40} \times 5^{20}$$

We can choose $$m^2$$ from any number of factors $$2$$ and $$5$$ as long as they are even (including 0), therefore the total number of combinations is $$21 \times 11 = 231$$.

Scroogey

science

[Daily puzzle] How many pairs?

south_korea_ln

I think those are the ones based on $$b^m \times b^n = b^{m \times n}$$

$$
(20^{10},1) (20^9,20^2) (20^8,20^4) (20^7,20^6) ... (20,20^{18}) (1,20^{20})
$$

But there are more based on $$b^n \times  c^n = (b \times c)^n$$ like

$$
(10^{10},2^{20}) (5^{10},4^{20}) (4^{10},5^{20}) (2^{10},10^{20})
$$
