• n = (a^2)/2
• n = (b^3)/3
• n = (c^5)/5

A bizarrely simple problem. Clearly n comprises powers of 2,3, and 5, and no others, to some degree.

If, say, 5n=c^5 then powers of 2,3,and 5 must all occur to a multiple of 5. So n must contain a power of 5 that is one less than a multiple of 5, that is, a number of the form (5k-1), that is also divisible by 2 and 3. Since such numbers are in the series {4,9,14,19,24,...}, 24 is the smallest possible candidate.

In like manner, the power of 2 is of the form (2k-1), and divisible by 3 and 5, so can't be less than 15, and the power of 3 is of the form (3k-1), and divisible by 2 and 5, so can't be less than 20.

The least n is then 2^15*3^20*5*24; 2n=(2^8)^2*(3^10)^2*(5^12)^2, a square; 3n=(2^5)^3*(3^7)^3*(5^12)^3, a cube, and 5n = (2^3)^5*(3^4)^5*(5^5)^5, a fifth power.