Can't solve this Number Theory problem.

Question

Define $A$ as the smallest integer for which $\sqrt{10 A}$ is an integer and $\sqrt[3]{6A}$ is an integer too. Get the number of Factors of A.

I think that I don't need to find the $A$ to solve this problem, there is definitely some pattern in Number Theory that I don't know.

And Also, the expected time to solve this problem is 3 minutes. I found this problem from an exam and it was among the easiest questions(there are 20 questions for "easy" category, each giving 2 points, this was 11th), and I expect it to have a simple solution.

Answer 1

Hint:

Let $n=A=2^a3^b5^cd$ where $(d,30)=1$

$$10n=2^{a+1}3^b5^{c+1}d$$

So, $a+1,b,c+1$ must be even and $d$ must be perfect square

$a+1=2e,b=2f,c+1=2g, d=h^2$

$$6n=2^{a+1}3^{b+1}5^cd=2^{2c}3^{2f+1}5^{2g-1}h^2$$

So, $3$ must divide $2e,2f+1,2g-1(1)$ and $h^2$ must be perfect cube $\ \ \ \ (2)$

$(1)\implies3|2e\iff3|e,e=3k$(sa)

$3|(b+1),b+1=3m$(say)

$3|(2g-1)\iff g=3n+2$

$(2)\implies h$ must be perfect cube $h=r^3$

So, $a=2e-1=2(3k)-1$

$b=3m-1$

$c=2g-1=2(3n+2)-1=6n+3$

Answer 2

You have to say that $A\gt0$ because if not a trivial answer is $A=0$. Well, we have $10A=x^2$ and $6A=y^3$ which implies the diophantine equation $$10y^3=6x^2$$ Among the solutions, the smallest $A$ corresponds to $(x,y)=(600,60)$ which gives $A=36000$

Now $36000=2^5\cdot3^2\cdot 5^3$ and it is known that there are $(5+1)(2+1)(3+1)=72$ factors.