One step ahead of you

To be divisible by 2000 ${ a }^{ a }{ b }^{ b }$ must be the product of at least one 2 and at least three 10's. Then let $b=10$ . It follows that $a=1$ .

$\ 2000 = 2^4 \times\ 5^3$ so at least one of a and b must contain a factor of 5 and the same goes goes for the factor of 2. $\therefore 10 | ab$ (10 is a factor of ab). $\ (a,b) = (10,1)$ gives the required solution.

If $a=10$ and $b=1$ then we get a solution of $ab = 10$ . There are only few pairs of natural numbers $(a,b)$ such that their product is strictly less than 10 and they are $(1,1), (1,2), \dots , (1,9), (2,2), (2,4)$ and $(3, 3)$ . Testing each of these shows that none of them satisfy $2000 | a^a \times b^b$ so therefore the least possible value of $ab$ is $\boxed{10}$

To be divisible by 2000, a^a * b^b must be multiple of one 2 and three 10's . As the both numbers to multiplied are the form of a^a , from this we can easily conclude that 10 must be one of a or b. Then taking a=10 and b=1( as 10^10 contains multiple of one 2 and three 10's , hence selection of b is easy ), we can conlude.