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Determine the number of positive integers a less than 12 such that the congruence \(ax\equiv 1\pmod{12}\) has a solution in x.

ant101  Apr 2, 2017
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By simple inspection, we have:

 

a=1   and  x=1

a=5   and  x=5

a=7  and  x=7

a=11 and x=11

Guest Apr 2, 2017

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