Natural number?

Some people consider 0 a member of N, but I'm assuming you don't or the problem's solution would be trivially 0.

If 0 is not a natural number, then the answer is a very, very large number. I've got it worked out here, but since I'm over 17, I'll let one of you youngsters figure it out.

Okay, now I'll answer:

The answer is of the form N= 2^x * 3^y * 5^z. We know the following facts:

2 divides x+1, y, and z.

3 divides x, y+1, and z

5 divides x,y, and z+1

That means we can use the chinese remainder theorem for x, y, and z, and get that:

x == 15 (mod 30)

y == 20 (mod 30)

z == 24 (mod 30)

So the smallest positive natural number that has this property is:

2^15 * 3^20 * 5^24

10

(2*N)^(1/2) is an integer if N is of the form 2*2^(2x) where x is a natural.

(3*N)^(1/3) is an integer if N is of the form 9*3^(3x) where x is a natural.

(5*N)^(1/5) is an integer if N is of the form 625*5^(5x) where x is a natural.

I'll let the youngsters figure it out.

2N is an even number. To have an integer square root, it must be divisible by 2^2 = 4, so N must be divisible by 4/2 = 2.

3N is a number divisible by 3. To have an integer cube root, it must also be divisible by 3^3 = 27, so N must be divisible by 27/3 = 9.

5N is a number divisible by 5. To have an integer fifth root, it must also be divisible by 5^5 = 3125, so N must be divisible by 3125/5 = 625.

Therefore, the lowest possible N is the least common multiple of 2, 9, and 625. They are all mutually prime, so the LCM is simply their product, 2*9*625 = 11250.

Thomas, I meant to give you a thumbs up, but the stupid mouse moved on me and it turned into a thumbs down. So consider that 2 thumbs up.

(2 * N)^(1/2)

2N^2

4N

N = 4/2 or 2

(3 * N) ^ (1/3)

3N^3

27N

N = 27/3

N = 9

(5 * N) ^ (1/5)

5N^5

3125N

N = 3125/5

N = 625