what is the smallest positive integer in the set $\{24x+60y+2000z \mid x,y,z \in \mathbb{Z}\}$! I cant understand how to do it please help me. Thanks in advance my question is  what is the smallest positive integer in the  set   $\{24x+60y+2000z \mid x,y,z \in \mathbb{Z}\}$! its options are given $2,4,6,24$
 A: As $(24,60)=12,$ 
$24a+60b=12(2a+5b)$ will be divisible by $12$  for integers $a,b$
Using Bézout's Lemma/identity, we can find integers $x,y$ such that $24x+60y=12$
In fact $60(1)+24(-2)=12$
Now, $(12,2000)=4,$ we can find integers $p,q$ such that $12p+2000q=4$
By actual division $2000=167\cdot12-4\implies  167\cdot12-2000=4$
$$\implies 167\{60(1)+24(-2)\}-2000=4$$
$$\implies 167\cdot60+24(-2\cdot167)+2000(-1)=4$$
This is the smallest as $24x+60y+2000z=4(6x+15y+500z)$ will always be divisible by $4$ for integers $x,y,z$
A: The set in question is a subgroup of $\mathbb{Z}$, and hence generated by a single element $c \in \mathbb{N}$, which is also the smallest positive integer in the set. This element must be a common divisor of 24, 60, and 2000 (since each of those elements is in this set). It is not too hard to see that $c$ must be the greatest common divisor of all these three numbers, which is 4.
A: $$24x+60y+2000z=4(6x+15y+500z)$$ Now, $$\gcd(6,15,500)=1\Rightarrow 6x+15y+500z=1$$ for some $x,y,z\in \mathbb{Z}$ by using the generalized Bezout's identity. Hence $4$ is the answer. 
In general the minimum positive integer value that the set $\displaystyle \left\{\sum_{k=1}^n a_kx_k\big|x_k\in\mathbb{Z},1\le k\le n\right\}$ can give is $\gcd(a_1,a_2,\cdots\ ,a_n)$. 
A: Bézout's theorem about G.C.D answers your question I think:
Definition: We say $d=\gcd(a,b,c)$ is the greatest common divisor of $a$,$b$ and $c$ if and only if the following holds:
1- $d>0$ 
2- $d|a$ , $d|b$ and $d|c$ 
3- If $e|a$ , $e|b$ and $e|c$ then $d|e$
Now we prove the existence of $\gcd(a,b,c)$ for any $a,b,c \in \mathbb{Z}$ using Bézout's theorem:
We claim that $d= \min\{ ax+by+cz>0: x,y,z \in \mathbb{Z} \}$:
It's clear that that $S=\{ ax+by+cz>0: x,y,z \in \mathbb{Z} \} \neq \emptyset$. Because either $a.1+b.0+c.0$ or $a.(-1)+b.0+c.0$ is positive and hence is in $S$. Therefore, $S$ is a non-empty subset of natural numbers and by using the well-ordering principle $S$ has a least positive element. set $d=\min(S)$. We claim that d satisfies all of the properties of $\gcd(a,b,c)$:
1) $d>0$ is trivial by the definition of $S$. 
2) We claim that $d|a$, the same technique could be used to prove that $d|b$ and $d|c$
Using Euclid's division algorithm(theorem) we know that $\exists q,r \in \mathbb{Z}$ such that $a=qd+r$ where $0 \leq r < d$
But $d \in S$ therefore $\exists x_0,y_0,z_0 \in \mathbb{Z}$ such that $d=a.x_0 + b.y_0 + c.z_0$. Therefore $a =q(a.x_0 + b.y_0 + c.z_0) + r \implies 0 \leq r = (1-qx_0)a + (-qy_0)b+(-qz_0)c \in S$. If $r>0$ then $r<d$ contradicts that $d = \min(S)$. Thefore $r=0$ and $a=qd$ which means $d|a$. You can similarly show $d|b$ and $d|c$.
3) if $e|a$ , $e|b$ and $e|c$ then $e|a.x_0+b.y_0+c.z_0, i.e. $e|d$
This proves that $d= \min\{ ax+by+cz>0: x,y,z \in \mathbb{Z} \}$ is equal to $\gcd(a,b,c)$. Therefore, all you need to do is to calculate $\gcd(a,b,c)$ and you can do that using Euclid's algorithm.
