as stated in the title - Find the smallest integer k such that 600k is a cube number. Thanks ever so much!
Hello,
The trick for this is to factor 600 :
$\displaystyle 600=4\times 150=8\times 3\times 5^2=2^3 \times 3\times 5^2$
But in a cube number, the prime decomposition is in the form $\displaystyle p_1^{3\alpha_1}\times p_2^{3\alpha_2}\times\dots$
So for $\displaystyle 2^3$, it's okay.
If 3 is in the prime decomposition of the cube number, then $\displaystyle 3^3$ has to. So keep a factor $\displaystyle 3^2$
Same reasoning for 5. Keep a factor 5.
$\displaystyle k=3^2\times 5=45$ is the smallest integer such that $\displaystyle 600k$ is a cube number.
This is a rather intuitive thing. I'm sorry if I'm not providing a formal proof :s
I think you mean positive integer k:
$\displaystyle 600k = 2^3 \times3 \times 5^2 \times k$
Now we see that 2 is already cubed. If we throw a couple of threes in k and one 5, we will get 3 and 5 cubed...
So choose $\displaystyle k = 3^2 \times 5 = 45$
$\displaystyle 600k = 2^3 \times3 \times 5^2 \times 3^2 \times 5 = 2^3 \times 3^3 \times 5^3 = (2 \times 3 \times 5)^3 = 30^3$
Can you see why this must be the smallest integer k?