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

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

implies that n is divisible by 2, 3, and 5.  The smallest n should only be divisible by these three factors or else it's not smallest.  So let n=(2^r)(3^s)(5^t)

The first equation implies a^2=2n=(2^(r+1))(3^s)(5^t), thus
2|r+1, 2|s, 2|t

Similarly, b^3=(2^r)(3^(s+1))(5^t) implies
3|r, 3|s+1, 3|t

and c^5=(2^r)(3^s)(5^(t+1)) implies
5|r, 5|s, 5|t+1

Combining them, we get
2|r+1, 3|r, 5|r
2|s, 3|s+1, 5|s
2|t, 3|t, 5|t+1

Smallest n corresponds to the smallest r, s, t.  And the smallest values are r=15, s=20, t=24

so n = (2^15)(3^20)(5^24)

that's pretty big.