A Power of Every Kind

Find a number, somewhere between $3$ and three thousand million million million (that is $3 \times 10^{21}$), that is at once a perfect square, a perfect cube, a perfect fourth power, a perfect fifth power and a perfect sixth power.

Solution

To be a square and a cube and a fourth, fifth and sixth power all together, a number must be a perfect $L$th power where $L$ is a common multiple of $2, 3, 4, 5$ and $6$. The least such $L$ is their lowest common multiple, $$\operatorname{lcm}(2,3,4,5,6) = 60,$$ so we want a sixtieth power. A sixtieth power is automatically all the rest, since $60$ is a multiple of each: for instance it is the square of a thirtieth power and the cube of a twentieth.

The smallest sixtieth power above $3$ is $$2^{60} = 1{,}152{,}921{,}504{,}606{,}846{,}976 \approx 1.15 \times 10^{18},$$ which sits comfortably below the stated ceiling. The next, $3^{60}$, is about $4.2 \times 10^{28}$, far beyond it. So $2^{60}$ is the one and only answer in range.