What is the smallest positive integer of the form $30x+6y+10z$? I am trying to find the smallest positive integer of the form
$30x+6y+10z$, where $(x,y,z)\in\mathbb{Z}$ 
However, I do not know where to start. Hints or answers are welcome.
Thanks!
 A: Hint: The integer has to be even (can you see why?). $6$ yes. Can you get $4$ or $2$? If you can generate $2$ you're done.
A: Since $\mathbf Z$ is a PID, it is the gcd of 30, 6 and 10 — 2.
A: Hint $\ $ The set $\rm\,S\,$ of integers of form $\rm\,a_1\,x_1 + \cdots + a_k x_k,\ x_i\in \mathbb Z,\,$ is closed under subtraction so, by this Lemma, every positive $\rm\,n\in S\,$ is divisible by $\rm\,d = $ least positive $\rm\in S.\,$  Now $\rm\,a_i\in S$ $\,\Rightarrow\,$ $\rm d\mid a_i,\,$ i.e. $\rm\,d\,$ is a common divisor of all $\rm\,a_i,\,$ necessarily greatest: $\rm\ c\mid a_i$ $\Rightarrow$ $\rm\,c\mid d = a_!\,x_1\!+\!\cdots\!+\!a_k x_k$ $\Rightarrow$ $\rm\,c\le d.$   
Remark $\ $ This is the general form of Bezout's Identity for the gcd.
A: See that $30 - 10 - 3\cdot 6 = 2 (= 2\cdot 10 - 3\cdot 6 = 2\cdot 6 - 10)$.
Also note that in general there always exists an integer linear combination of $\{x,y,z\}$ such that
$$\lambda_1 x+ \lambda_2 y + \lambda_3 z = \gcd(x,y,z)$$
Since $\gcd(30,10,6) = 2 \ne 1$ we know that $2$ is in fact the smallest possible integer. $\gcd(x,y,z)$ will always be the least positive integer of that form in general, this is known as Bézout's lemma.
