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How many positive integers n satisfy 127 = 3 (mod n)? n = 1 is allowed.

 Nov 26, 2020
 #1
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n = 4,  31,  62,  124

 Nov 26, 2020
 #2
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Subracting \(3\) from both sides gives \(124=0 \pmod n\). This means that \(n\) is a factor of \(124\). Since \(124=2^2 \cdot 31^1\), the amount of factors \(124\) has is \((2+1)(1+1)=6\), which should be the answer.

 Nov 26, 2020

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