GCD of pairs of integers So I think I am just psyching myself out right now and this is way to easy but I am running on no sleep in the past few days so forgive me please. The question is what are the greatest common divisors of these pairs of integers. 
1) $2^3\cdot3^2, 2^2\cdot3^3$
2) $2^3\cdot3^2, 5^2\cdot7^3$
3) $2^3\cdot3^2, 0$
Would the answer for 1 be $2^2\cdot3^2$. Number 2) $1$ Number 3) $2^3\cdot3^2$? 
 A: Yes, you are correct.
Since your numbers already are factorized into primes, you simply take as many common factors as possible, as you have done. The first pair shares $2^2 \cdot 3^2$.
The second pair share no prime powers, so we say they are relatively prime and hence $\gcd = 1$.
Because every number divides 0 we have for the third pair $\gcd(2^3 \cdot 3^2, 0) = 2^3 \cdot 3^2$. A more detailed description of this is for example available here.
A: Given a positive integer $n$, it is always possible to express it as $$n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$$ for some primes $p_1,\dots,p_k$ and nonnegative integers $\alpha_1,\dots,\alpha_k$. So if we have two numbers $n,m$, we can write $$n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k} \\ m=p_1^{\beta_1}p_2^{\beta_2}\dots p_k^{\beta_k} $$ where $p_1,\dots,p_k$ is the union of the sets of prime factors of $n$ and $m$ and some of the $\alpha_i$ and $\beta_i$ are possibly $0$. Then $$\gcd(n,m)=p_1^{\min\{\alpha_1,\beta_1\}}p_2^{\min\{\alpha_2,\beta_2\}}\dots p_k^{\min\{\alpha_k,\beta_k\}}$$
So here, the first pair is $2^3\times 3^2$ and $2^2\times 3^3$, so their gcd is $2^{\min\{3,2\}}\times 3^{\min\{2,3\}}=2^2\times3^2$. Similarly for the second one. So yes, you were right.
To get the last one, just recall that every natural number divides 0.
Incidentally, it is also true that $$\text{lcm}(n,m)=p_1^{\max\{\alpha_1,\beta_1\}}p_2^{\max\{\alpha_2,\beta_2\}}\dots p_k^{\max\{\alpha_k,\beta_k\}}$$
