# Quotient

Find quotient before the bracket - the largest divisor
51 a + 34 b + 68
121y-99z-33

Result

d1 =  17
d2 =  11

#### Solution:

$51=3 \cdot 17 \ \\ 34=2 \cdot 17 \ \\ 68=2^2 \cdot 17 \ \\ \text{GCD}(51, 34, 68)=17 \ \\ \ \\ d_{1}=GCD(51,34,68)=17 \ \\ \ \\ 51a+34b+68=17(3a+2b+4)$
$121=11^2 \ \\ 99=3^2 \cdot 11 \ \\ 33=3 \cdot 11 \ \\ \text{GCD}(121, 99, 33)=11 \ \\ \ \\ d_{2}=GCD(121,99,33)=11 \ \\ \ \\ 121y-99z+33=11(11y-9z+3)$

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