Minimize $152207x-81103y$ over the positive integers. 
Minimize the expression $152207x-81103y$ over the positive integers,
  given $x,y\in\mathbb{Z}.$

So the book takes me through modular arithmetic and how to find $\text{gcd}(a,b)$ in order to solve diophantine equations. Then this question pops up in the same chapter.
I know how calculate using modular arithmetic, I know how to find $\text{gcd}(a,b)$ and solve diophantine equations but I don't know how to bunch them up together in order to solve this.
How should I think?
 A: Since $\gcd (152207,81103)=1111$ it is the same as minimum of $$1111(137x-73y)$$
Since $137x-73y=1$ is solvable (say $x=8$ and $y=15$) the answer is $1111$.
A: Note $c=ax+by$ has integer solution if and only if $gcd(a,b)|c$ and if this exists then infinite no of integer solutions can be obtained from 1 solution by 
$x=x_{0}+\frac{b}{d}k$ and $y=y_{0}-\frac{a}{d}k$ 
where $d$ is the gcd.
And $x_{0},y_{0}$ are one solution which can be obtained from euclid's algorithm. (As will jagy has mentioned) 
And also the defination of $gcd(a,b)$ is least positive value of the $ax+by=d$  where $x,y$ any integer.
A: $$  \gcd( 152207, 81103 ) = ???    $$ 
$$ \frac{ 152207 }{ 81103 } = 1 +  \frac{ 71104 }{ 81103 } $$
$$ \frac{ 81103 }{ 71104 } = 1 +  \frac{ 9999 }{ 71104 } $$
$$ \frac{ 71104 }{ 9999 } = 7 +  \frac{ 1111 }{ 9999 } $$
$$ \frac{ 9999 }{ 1111 } = 9 +  \frac{ 0 }{ 1111 } $$ 
 Simple continued fraction tableau:
$$ 
 \begin{array}{cccccccccc}
 & & 1 & & 1 & & 7 & & 9 & \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 1 }{ 1 }   & &   \frac{ 2 }{ 1 }   & &   \frac{ 15 }{ 8 }   & &   \frac{ 137 }{ 73 }  
 \end{array}
 $$
$$  $$
$$ 137 \cdot 8 - 73 \cdot 15 = 1 $$ 
$$  \gcd( 152207, 81103 ) = 1111  $$
$$ 152207 \cdot 8 - 81103 \cdot 15 = 1111 $$ 
