The least positive integer that is divisible by 2,3,4 and 5, and is

It is a 4th power of 2



Posted from my mobile device

It's not a 4th power:

\(\\
60^{30} = (2^2 \times 3 \times 5)^{30} = 2^{60} 3^{30} 5^{30}\\
\)

If this number were a 4th power, every exponent in its prime factorization would need to be a multiple of 4, but that's not the case. The number is a square, a cube, and a 5th power, because the prime factorization exponents are multiples of 2, 3 and 5, so it's close to meeting the requirements but isn't quite right.

The smallest number that works here is \(30^{60}\). That number is clearly a power of 2, 3, 4 and 5, because the exponent is a multiple of 2, 3, 4 and 5. It's also clearly divisible by 2, 3 and 5 because the base 30 is divisible by those numbers, but it will also certainly be a multiple of 4 = 2^2, because we'll end up finding 2^60 in its prime factorization. So the answer is 30 + 60 = 90, since there's no smaller possibility.

Thanks IanStewart for clarifying my doubt and this is exactly what I was going to ask to Kinshook

Its a perfect square because: \((60^{15})^2\)
Its a cube because: \((60^{10})^3\)
Its a 5th power because: \((60^{6})^5\)

But it is not a 4th power because 30 is not a multiple of 4. However, if we interchange the values of the base and the exponent then that makes sense i.e \(30^{60}\) as it satisfies all the given conditions.