# Quotient

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

Result

d1 =  17
d2 =  11

#### Solution:

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

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