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Question Number 121834 by bemath last updated on 12/Nov/20

Answered by liberty last updated on 12/Nov/20

$$\mathrm{we}\mathrm{have}\gcd (93,27)=3\mathrm{and}6\mathrm{divides}\mathrm{by}3$$$$\mathrm{therefore}\mathrm{there}\mathrm{exist}\mathrm{solutions}.$$$$\mathrm{we}\mathrm{have}93=3\times 27+12\wedge 27=2\times 12+3$$$$12=4\times 3+0$$$$3=1\times 27-2\times 12$$$$=1\times 27-2(1\times 93-3\times 27)$$$$=7\times 27-2\times 93$$$$\mathrm{thus}-2\times 93+7\times 27=3$$$$\mathrm{so}-4\times 93-(-14)\times 27=6$$$$\mathrm{we}\mathrm{get}\mathrm{general}\mathrm{solutions}\mathrm{is}\{\begin{array}{l}\mathrm{x}=-4-9\mathrm{k}\\ \mathrm{y}=-14-31\mathrm{k}\end{array}$$$$\mathrm{for}\mathrm{all}\mathrm{k}\in \mathbb{Z}$$$$$$

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