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⋅17 34=2⋅17 68=22⋅17 GCD(51,34,68)=17  d1=GCD(51,34,68)=17  51a+34b+68=17(3a+2b+4)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=112 99=32⋅11 33=3⋅11 GCD(121,99,33)=11  d2=GCD(121,99,33)=11  121y−99z+33=11(11y−9z+3)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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