Quotient

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

Correct result:

d₁ =  17
d₂ =  11

Solution:

$$\begin{gathered} 51=3\cdot 17 \\ 34=2\cdot 17 \\ 68=2^2\cdot 17 \\ \text{GCD}(51,34,68)=17 \\ {d}_1=GCD(51,34,68)=17 \\ 51a+34b+68=17(3a+2b+4) \end{gathered}$$

$$\begin{gathered} 121=11^2 \\ 99=3^2\cdot 11 \\ 33=3\cdot 11 \\ \text{GCD}(121,99,33)=11 \\ {d}_2=GCD(121,99,33)=11 \\ 121y-99z+33=11(11y-9z+3) \end{gathered}$$

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