Given that \(N=a^3b^4c^5\) where \(a, b\) and \(c\) are distinct prime numbers, what is the smallest number with which

multiple of square, cube and 5th power; simply take the LCM of power required = 2x3x5 = 30.
We need 30 in each power to make it follow what is asked.
It'll be a^30 b^30 c^30
Hence, C.

To be a perfect square, a perfect cube as well as a perfect fifth power, such number should be in power of this expression which is divisible by 2,3 and 5 , In simpler terms, we are required to find LCM of 2, 3 & 5 which is 30... as it should be perfect square AND cube AND Fifth.

So, 30 should be the power of each expression a, b, & c

OA: E (30-3) (30-4) (30-5)

The power exponents of a, b and c have to be divisible by 2, 3, and 5 for N to be a perfect square, perfect cube, and perfect 5th power.
Therefore, we need to find the LCM of 2, 3, and 5.
2 = 1 * 2
3 = 1 * 3
5 = 1 * 5
LCM = 1 * 2 * 3 * 5
$$Therefore,\ the\ smallest\ integer\ N=a^{30}b^{30}c^{30}$$
$$N=a^{\left(30-3\right)}b^{\left(30-4\right)}c^{\left(30-5\right)}$$
$$N=a^{27}b^{26}c^{25}\ \ \ \ \ \ \ \ \ Answer\ =\ option\ E$$

Thank you for the solution. That was really meaningful.