Question

part g - division show work for partial credit /2 pts each
#13 (mod 17): calculate 7÷16
#14 (mod 14): calculate 1÷5

Explanation:

Step1: Recall modular - division as modular - multiplication

In modular arithmetic, dividing \(a\) by \(b\) modulo \(m\) is equivalent to multiplying \(a\) by the multiplicative inverse of \(b\) modulo \(m\). That is, \(a\div b\equiv a\times b^{-1}\pmod{m}\), where \(b\times b^{-1}\equiv1\pmod{m}\).

Step2: Find the multiplicative inverse for #13

We want to find the multiplicative inverse of \(16\) modulo \(17\). Since \(16\equiv - 1\pmod{17}\), and \((-1)\times(-1) = 1\), the multiplicative inverse of \(16\) modulo \(17\) is \(16\) itself. Then \(7\div16\equiv7\times16\pmod{17}\). Calculate \(7\times16 = 112\), and \(112\div17 = 6\) with a remainder of \(10\). So \(7\div16\equiv10\pmod{17}\).

Step3: Find the multiplicative inverse for #14

We need to find the multiplicative inverse of \(5\) modulo \(14\). We want to solve the congruence \(5x\equiv1\pmod{14}\). We can use the Extended - Euclidean algorithm:
We know that we want to find \(x\) and \(y\) such that \(5x+14y = 1\).
Apply the Euclidean algorithm: \(14 = 2\times5+4\), \(5 = 1\times4 + 1\).
Then \(1=5-4\), and since \(4 = 14 - 2\times5\), we substitute \(4\) into the equation: \(1=5-(14 - 2\times5)=3\times5-14\). So \(x = 3\) is the multiplicative inverse of \(5\) modulo \(14\). Then \(1\div5\equiv1\times3\pmod{14}\equiv3\pmod{14}\).

Answer:

#13: \(10\)
#14: \(3\)