The smallest positive integer $$n$$ for which $$18900 \times n$$ is a perfect cube is

Given

$$18900\times n$$ is a perfect cube.

 Step 1: Prime factorize 18900

$$18900=189\times100$$

$$=3^3\times7\times2^2\times5^2$$

Therefore,

$$18900=2^2\times3^3\times5^2\times7.$$

 Step 2: Make the powers multiples of 3

For a number to be a perfect cube, the power of every prime factor must be a multiple of $$3$$.

We have $$18900=2^2\times3^3\times5^2\times7^1.$$

Therefore, we need to multiply by: $$2^1$$ to make the power of $$2$$ equal to $$3$$,

$$5^1$$ to make the power of $$5$$ equal to $$3$$,

and $$7^2$$ to make the power of $$7$$ equal to $$3$$.

Hence, $$n=2\times5\times7^2.$$

$$=2\times5\times49.$$

$$=490.$$

Final answer $$\boxed{490}$$