Modular Arithmetic - SS1 Mathematics Past Questions and Answers - page 1
What is the time 11 hours after 6 o′clock (6:00 am)?
5:00 pm
6:00 pm
10:00 pm
4:00 pm
Convert 34 and −34 to their congruent values in mod 9
(a) \(34\ (mod\ 9)\ \equiv x\)
\[\frac{34}{9} = 3\ remainder\ \mathbf{7}\]
Thus, \(34\ (mod\ 9)\ \equiv \mathbf{7}\)
(b) \(- 34\ (mod\ 9)\ \equiv x\)
\[- 34 = x\ (mod\ 9)\ \ \]
\[- 34 = q.9 + x\]
\[- 34 = \ - 4(9) + x\]
\[- 34 = - 36 + x\]
\[- 34 + 36 = x\]
\[x = \mathbf{2}\]
Evaluate the following: (i) \(5\ \oplus \ 2\ (mod\ 3)\) (ii) \(11 \ominus 23\ (mod\ 3)\) (iii) \(4\ \otimes 16\ (mod\ 4)\)
Thus, \(- 34\ (mod\ 9)\ \equiv \mathbf{2}\)
-
(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, \(4\ \otimes 16\ (mod\ 4)\ \equiv \mathbf{0}\)
What is the time \(6\ hours\) after \(10\ o’clock\) at night (\(10:00\ pm\))?
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.