Find the smallest number $n > 1$ such that $n = a^2 = b^3 = c^3$ Find the smallest number $n > 1$ such that 
\begin{equation}
n = a^2 = b^3 = c^5
\end{equation}
for three integers $a, b$ and $c$. 
I thought about finding $\mathrm{lcm}(a^2,b^3)$ and $\mathrm{lcm}(b^3, c^5)$ by using the formula
\begin{equation}
\mathrm{lcm}(x,y) = \frac{x\cdot y}{\gcd(x,y)}
\end{equation}
and compare them to find $n$, but I'm struggling to find an expression for $\gcd(a^2,b^3)$ and $\gcd(b^3, c^5)$ that makes any sense. 
I've tried to make expressions of $a$, $b$ and $c$, only to come up with this
\begin{align*}
a = b\sqrt{b} \quad &\Rightarrow \gcd(a, b) = b  &\Rightarrow \mathrm{lcm}(a,b) = a\\
a = c^2 \sqrt{c}\quad  &\Rightarrow \gcd(a,c) = c &\Rightarrow \mathrm{lcm}(a,c) = a\\
b = c^2 \frac{\sqrt{c}}{\sqrt{b}} \quad &\Rightarrow \gcd(b,c) = c &\Rightarrow \mathrm{lcm}(b,c) = b
\end{align*}
and
\begin{align*}
\gcd(a^2, b^3) = b^3 &\Rightarrow \mathrm{lcm}(a^2,b^3) = a^2 \\
\gcd(a^2, c^5) = c^5 &\Rightarrow \mathrm{lcm}(a^2,c^3) = a^2 \\
\gcd(b^3, c^5) = c^5 &\Rightarrow \mathrm{lcm}(b^3, c^5) = b^5 \\
\end{align*}
which doesn't get me anywhere. 
Any suggestions to how I should tackle this problem? 

Let $n=p_1^{q_1}p_2^{q_2}...p_n^{q_n}$ be the prime factorization of $n$. Then we see that for $n$ to be a square, a cube, and a fifth power, we must have $2|{q_1}, 2|{q_2}, ... 2|{q_n}$, $3|{q_1}, 3|{q_2}, ... 3|{q_n}$, and $5|{q_1}, 5|{q_2}, ... 5|{q_n}$. To find the smallest $n$, we can let $n$ be of the form $p_1^{q_1}$ and choose $p_1=2$ as the smallest prime. Now we need to solve for $q_1$. Since we have $2|q_1$, $3|q_1$, and $5|q_1$, we see that $q_1=lcm(2,3,5)=30$. Then we have that $n=2^{30}$.
