Modular Arithmetic - SS1 Mathematics Past Questions and Answers - page 1

(a) \(34\ (mod\ 9)\ \equiv x\)

\[\frac{34}{9} = 3\ remainder\ \mathbf{7}\]

Thus, \(34\ (mod\ 9)\ \equiv \mathbf{7}\)

(b) \(- 34\ (mod\ 9)\ \equiv x\)

\[- 34 = x\ (mod\ 9)\ \ \]

\[- 34 = q.9 + x\]

\[- 34 = \ - 4(9) + x\]

\[- 34 = - 36 + x\]

\[- 34 + 36 = x\]

\[x = \mathbf{2}\]

Thus, \(- 34\ (mod\ 9)\ \equiv \mathbf{2}\)

1. (i) \(5\ \oplus \ 2\ (mod\ 3)\ \ \ = 7\ (mod\ 3)\)

\(\frac{7}{3} = 2\ remainder\ \mathbf{1}\)

Thus, \(5\ \oplus \ 2\ (mod\ 3)\ \ \ \equiv \mathbf{1}\)

(ii) \(11 \ominus 23\ (mod\ 3) = \ - 12(mod\ 3)\)

\[- 12 = q.3 + x\]

\[- 12 = - 4(3) + x\]

\[- 12 = - 12 + x\]

\(- 12 + 12 = x\)

\[x = \mathbf{0}\]

Thus, \(11 \ominus 23\ (mod\ 3)\ \equiv \mathbf{0}\)

(iii) \(4\ \otimes 16\ (mod\ 4) = 64\ (mod\ 4)\)

\(\frac{64}{4} = 16\ remainder\ \mathbf{0}\)

Thus, \(4\ \otimes 16\ (mod\ 4)\ \equiv \mathbf{0}\)

Assuming a 12-hour clock, that is modulo 12

\(6\ \oplus 10\ (mod\ 12)\ \ \ \ \ = 16\ (mod\ 12)\ \ \ \ \ \ = \ \frac{16}{12}\ = 1\ remainder\ \mathbf{4}\)

Thus, \(6 \oplus 10\ (mod\ 12)\ \equiv \mathbf{4}\), meaning the time will be 4 o’clock or 4:00 am the next day.