Using Euclidean Algorithm in RSA? it leverages multiplication and subtraction, which humans are fairly good at, to make fractions like 15996751/3870378 reducible. Also useful in scaling equations down to their simplest integer representation in one step, given with extra integers, GCD(C,GCD(A,B)) is equivalent to GCD(A,B,C). It's been around for over two thousand years and mathematicians have bent it to many purposes , but just the first bit should be justification.
it leverages multiplication and subtraction, which humans are fairly good at, to make fractions like 15996751/3870378 reducible. Also useful in scaling equations down to their simplest integer representation in one step, given with extra integers, GCD(C,GCD(A,B)) is equivalent to GCD(A,B,C). It's been around for over two thousand years and mathematicians have bent it to many purposes , but just the first bit should be justification.
ind a and b? 

Here is the continued fraction way of writing this. It is, of course, equivalent to the "back-substitution" way one often sees. The convergents can bbe written in a single row of fractions, as below. 
$$  \gcd( 1680, 71 ) = ???    $$  
$$ \frac{ 1680 }{ 71 } = 23 +  \frac{ 47 }{ 71 } $$ 
 $$ \frac{ 71 }{ 47 } = 1 +  \frac{ 24 }{ 47 } $$ 
 $$ \frac{ 47 }{ 24 } = 1 +  \frac{ 23 }{ 24 } $$ 
 $$ \frac{ 24 }{ 23 } = 1 +  \frac{ 1 }{ 23 } $$ 
 $$ \frac{ 23 }{ 1 } = 23 +  \frac{ 0 }{ 1 } $$ 
 Simple continued fraction tableau:
 $$ 
 \begin{array}{cccccccccccc}
 & & 23 & & 1 & & 1 & & 1 & & 23 & \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 23 }{ 1 }   & &   \frac{ 24 }{ 1 }   & &   \frac{ 47 }{ 2 }   & &   \frac{ 71 }{ 3 }   & &   \frac{ 1680 }{ 71 }  
 \end{array}
 $$ 
 $$  $$ 
 $$ 
 \begin{array}{ccc}
  \frac{ 1 }{ 0 }   &     \mbox{digit}  &  23  \\  
  \frac{ 23 }{ 1 }   &     \mbox{digit}  &  1  \\  
  \frac{ 24 }{ 1 }   &     \mbox{digit}  &  1  \\  
  \frac{ 47 }{ 2 }   &     \mbox{digit}  &  1  \\  
  \frac{ 71 }{ 3 }   &     \mbox{digit}  &  23  \\  
  \frac{ 1680 }{ 71 }   &     \mbox{digit}  &  0  \\  
 \end{array}
 $$ 
$$ 1680 \cdot 3 - 71 \cdot 71 = -1 $$ 
