Find the smallest integer greater than 1 that has a square root, a cube root, a fourth root, a fifth root, a sixth root, a seventh root, an eighth root, a ninth root, and a tenth root, all perfect.
I think it's 2^x
where x = lcm(2,3,4,...,10).
By the way, where did you get these problems?
EDIT: I might have misinterpreted this problem. Thinking of perfect powers, I interpreted that the square root through tenth root are all integer. But if we further require that they be perfect numbers then I have no idea what the solution looks like.
Suppose $\displaystyle n$ is a natural number with integer $\displaystyle 2$-dn, $\displaystyle 3$-rd, ... $\displaystyle k$-th roots. Now consider the prime decomposition of $\displaystyle n$ :
$\displaystyle n=\prod_{i=1}^m p_i^{\alpha_i}$
Clearly for each $\displaystyle i$ we must have $\displaystyle \alpha_i$ is divisible by $\displaystyle 2, 3, .., k$. Hence the smallest natural with the given property is:
$\displaystyle n=2^{\text{lcm}(2,3,..,k)}$
because this number has the required property and any other number with this property must be the product of powers of primes each such power also having the property etc.
CB