Question

Suppose that a and b are integers,\({\bf{a}} \equiv {\bf{11}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right),{\rm{ }}{\bf{and}}{\rm{ }}{\bf{b}} \equiv {\bf{3}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right)\). Find the integer c with\({\bf{0}} \le {\bf{c}} \le {\bf{18}}\)such that

a)\({\bf{c}} \equiv {\bf{13a}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

b)\({\bf{c}} \equiv {\bf{8b}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

c)\({\bf{c}} \equiv {\bf{a}} - {\bf{b}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

d)\({\bf{c}} \equiv {\bf{7a}}{\rm{ }} + {\rm{ }}{\bf{3b}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

e)\({\bf{c}} \equiv {\bf{2}}{{\bf{a}}^{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{3}}{{\bf{b}}^{{\bf{2}}{\rm{ }}}}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

f )\({\bf{c}} \equiv {{\bf{a}}^{{\bf{3}}{\rm{ }}}} + {\rm{ }}{\bf{4}}{{\bf{b}}^{\bf{3}}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

Short answer

1. 10
2. 5
3. 8
4. 10
5. 3
6. 14

Step-by-step solution

Concept of Division Algorithm 

• Let\(a\)be an integer and\(d\)be a positive integer.
• Then there are unique Integers\(q\)and\(T\)with\(0 \le \;r < d\)such that\(a = dq + r\).
• \(q\)is called the quotient and\(T\)is called the remainder.
• \(\begin{array}{*{20}{l}}{q = a{\rm{ div }}d}\\{\;r = a{\rm{ mod }}d}\end{array}\)
• Theorem 5: Let m be a positive integer. If

\(a \equiv b\left( {{\rm{mod }}m} \right){\rm{ and }}c \equiv d\left( {{\rm{mod }}m} \right),{\rm{ then }}a + c \equiv b + d\left( {{\rm{mod }}m} \right){\rm{ and }}ac \equiv bd{\rm{ }}\left( {{\rm{mod }}m} \right)\)

(a) Finding the integer c when \({\bf{c}} \equiv {\bf{13a}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

\(\begin{array}{l}a \equiv 11\;(\bmod \;19)\\b \equiv 3\;(\bmod \;19)\\0 \le c \le 18\end{array}\)

Use theorem 5:

\(\begin{array}{c}c \equiv 13a\;(\bmod \;19)\\ = 13 \cdot 11\;(\bmod \;19)\\ = 143\;(\bmod \;19)\\ = 10\;(\bmod \;19)\end{array}\)

We then obtain\(c = 10\;{\rm{with}}\;0 \le c \le 18\)

Hence the integer is 10.

 (b) Finding the integer c when \({\bf{c}} \equiv {\bf{8b}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

\(\begin{array}{l}a \equiv 11\;(\bmod \;19)\\b \equiv 3\;(\bmod \;19)\\0 \le c \le 18\end{array}\)

Use theorem 5:

\(\begin{array}{c}c \equiv 8b\;(\bmod \;19)\\ = 8 \cdot 3\;(\bmod \;19)\\ = 24\;(\bmod \;19)\\ = 5\;(\bmod \;19)\end{array}\)

We then obtain\(c = 5\;{\rm{with}}\;0 \le c \le 18\)

Hence the integer is 5.

(c) Finding the integer c when \({\bf{c}} \equiv {\bf{a}} - {\bf{b}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

\(\begin{array}{l}a \equiv 11\;(\bmod \;19)\\b \equiv 3\;(\bmod \;19)\\0 \le c \le 18\end{array}\)

Use theorem 5:

\(\begin{array}{c}c \equiv a - b\;(\bmod \;19)\\ = 11 - 3\;(\bmod \;19)\\ = 8\;(\bmod \;19)\end{array}\)

We then obtain\(c = 8\;{\rm{with}}\;0 \le c \le 18\)

Hence the integer is 8.

(d) Finding the integer c when \({\bf{c}} \equiv {\bf{7a}}{\rm{ }} + {\rm{ }}{\bf{3b}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

\(\begin{array}{l}a \equiv 11\;(\bmod \;19)\\b \equiv 3\;(\bmod \;19)\\0 \le c \le 18\end{array}\)

Use theorem 5:

\(\begin{array}{c}c \equiv 7a + 3b\;(\bmod \;19)\\ = 7 \cdot 11 + 3 \cdot 3\;(\bmod \;19)\\ = 77 + 9\;(\bmod \;19)\\ = 86\;(\bmod \;19)\\ = 10\;(\bmod \;19)\;\end{array}\)

We then obtain\(c = 10\;{\rm{with}}\;0 \le c \le 18\)

Hence the integer is 10.

(e) Finding the integer c when \({\bf{c}} \equiv {\bf{2}}{{\bf{a}}^{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{3}}{{\bf{b}}^{{\bf{2}}{\rm{ }}}}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

\(\begin{array}{l}a \equiv 11\;(\bmod \;19)\\b \equiv 3\;(\bmod \;19)\\0 \le c \le 18\end{array}\)

Use theorem 5:

\(\begin{array}{c}c \equiv 2{a^2} + 3{b^2}\;(\bmod \;19)\\ = 2 \cdot {11^2} + 3 \cdot {3^2}\;(\bmod \;19)\\ = 2 \cdot 121 + 3 \cdot 9\;(\bmod \;19)\\ = 242 + 27\;(\bmod \;19)\\ = 3\;(\bmod \;19)\end{array}\)

We then obtain\(c = 3\;{\rm{with}}\;0 \le c \le 18\)

Hence the integer is 3.

(f) Finding the integer c when \({\bf{c}} \equiv {{\bf{a}}^{{\bf{3}}{\rm{ }}}} + {\rm{ }}{\bf{4}}{{\bf{b}}^{\bf{3}}}{\rm{ }}\left( {{\bf{mod}}{\rm{ }}{\bf{19}}} \right).\)

\(\begin{array}{l}a \equiv 11\;(\bmod \;19)\\b \equiv 3\;(\bmod \;19)\\0 \le c \le 18\end{array}\)

Use theorem 5:

\(\begin{array}{c}c \equiv {a^3} + 4{b^3}\;(\bmod \;19)\\ = {11^3} + 4 \cdot {3^3}\;(\bmod \;19)\\ = 1331 + 4 \cdot 27\;(\bmod \;19)\\ = 1331 + 108\;(\bmod \;19)\\ = 1439\;(\bmod \;19)\\ = 14\;(\bmod \;19)\end{array}\)

We then obtain\(c = 14\;{\rm{with}}\;0 \le c \le 18\)

Hence the integer is 14.