 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/6/a/i/solution

3231871365134 = 1 \cdot 187 + 136, = 1 \cdot 136 + 51, = 2 \cdot 51 + 34, = 1 \cdot 34 + 17, = 2 \cdot 17 + 0.

The last nonzero remainder is

17

, so

g cd (323, 187) = 17.

As a check,

323 = 19 \cdot 17

and

187 = 11 \cdot 17

17 = - 4 \cdot 323 + 7 \cdot 187

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