What is the smallest positive integer that is a perfect square when divided by three and it is a perfect cube when divided by two?

Answer 1

648.

648/3=216 (216)^(1/3)=6
648/2=324 (324)^(1/2)=18

^^^^^ Actually, that answers a different question: "What is a number that when divided by three is a perfect cube and when divided by two is a perfect square?"

To solve the original question:

You want a number that when divided by three is a perfect square. That means it can be written as 3n2 for some integer n. You also want a number that when divided by two is a perfect cube. That means it can be written as 2m3 for some other integer m. These are two ways of writing the same number so

3n2 = 2m3

This is equivalent to

n2 = (2/3)m3

In order for the right side to be a perfect square, it must be a multiple of an even power of 2. Note that this is satisfied if m is a multiple of an odd power of 2. The right side must also be a multiple of an even power of 3. This is satisfied if m is a multiple of an odd power of 3. That is, m needs to be of the form:

m = 22k+132j+1 for j, k non-negative integers.

(As an aside, notice that three possible values of m are 6, 24, and 54.)

Going back to the original problem, the smallest number that satisfies the conditions comes from m = 6:

2 x 63 = 432. (Note that 432 / 3 = 144 = 122).

But there are infinitely many numbers that work. For example

2 x 243 = 27648. (Note that 27648 / 3 = 962).