Question

part f - multiplicative inverse show work for partial credit/2 pts each. #11 (mod 14): inverse of 6? #12 (mod 15): inverse of 8?

Explanation:

Step1: Recall modular - inverse formula

We want to find an integer $x$ such that $ax\equiv 1\pmod{m}$, where for #11, $a = 6$ and $m = 14$, and for #12, $a = 8$ and $m = 15$. This is equivalent to finding $x$ and $y$ such that $ax+my = 1$ by the Extended - Euclidean algorithm. But for small values, we can also try values of $x$ from $0$ to $m - 1$.

Step2: Solve for #11

We need to find $x$ such that $6x\equiv 1\pmod{14}$. We check $x = 0,1,\cdots,13$.
$6\times0\equiv0\pmod{14}$, $6\times1\equiv6\pmod{14}$, $6\times2 = 12\equiv12\pmod{14}$, $6\times3=18\equiv4\pmod{14}$, $6\times4 = 24\equiv10\pmod{14}$, $6\times5=30\equiv2\pmod{14}$, $6\times6 = 36\equiv8\pmod{14}$, $6\times7=42\equiv0\pmod{14}$, $6\times8 = 48\equiv6\pmod{14}$, $6\times9=54\equiv12\pmod{14}$, $6\times10 = 60\equiv4\pmod{14}$, $6\times11=66\equiv10\pmod{14}$, $6\times12 = 72\equiv2\pmod{14}$, $6\times13=78\equiv8\pmod{14}$. Since there is no $x$ in the set $\{0,1,\cdots,13\}$ such that $6x\equiv 1\pmod{14}$, $6$ has no multiplicative inverse modulo $14$ because $\gcd(6,14)=2
eq1$.

Step3: Solve for #12

We need to find $x$ such that $8x\equiv 1\pmod{15}$. We check $x = 0,1,\cdots,14$.
$8\times0\equiv0\pmod{15}$, $8\times1\equiv8\pmod{15}$, $8\times2 = 16\equiv1\pmod{15}$.

Answer:

#11: No multiplicative inverse
#12: 2