Question

Find the integer a such that

a) \({\bf{a}} = {\bf{43}}\left( {{\bf{mod}}{\rm{ }}{\bf{23}}} \right)\) and \( - 22 \le a \le 0\).

b) \({\bf{a}} = {\bf{17}}\left( {{\bf{mod}}{\rm{ }}{\bf{29}}} \right)\) and \( - 14 \le a \le 14\).

c) \({\bf{a}} = - {\bf{11}}\;\left( {{\bf{mod}}{\rm{ }}{\bf{21}}} \right)\) and \(90 \le a \le 110\).

Short answer

1. Since -3 is between -22 and 0:

\(a = - 3\)

b. Since -12 is between -14 and 14:

\(a = - 12\)

c. Since 94 is between 90 and 110:

\(a = 94\)

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.
• \(\begin{array}{*{20}{l}}{q = a{\rm{ div }}d}\\{\;r = a{\rm{ mod }}d}\end{array}\)

Step 2(a): Finding the integer a when \({\bf{a}} = {\bf{43}}\left( {{\bf{mod}}{\rm{ }}{\bf{23}}} \right)\) and \( - 22 \le a \le 0\).

\(a \equiv 43\;(\,\bmod \;\,23)\)

We can find \(a\) such that \( - 22 \le a \le 0\) by consecutively subtracting 23 from 43 until

we obtain a value between -22 and 0.

\(\begin{array}{c}a \equiv 43\;(\,\bmod \,23)\;\;\;\;\\\; \equiv 43 - 23\;(\,\bmod \;\,23)\\ \equiv 20\;(\,\bmod \,\;23)\\ \equiv 20 - 23\;(\,\bmod \,23)\\ \equiv - 3\;(\,\bmod \,\;23)\end{array}\)

Since -3 is between -22 and 0:

\(a = - 3\)

Hence the integer a is -3

Step 3(b): Finding the integer a when \({\bf{a}} = {\bf{17}}\left( {{\bf{mod}}{\rm{ }}{\bf{29}}} \right)\) and \( - 14 \le a \le 14\).

\(a \equiv 17\;(\,\bmod \,\;29)\)

We can a find \(a\) such that \( - 14 \le a \le 14\)by consecutively subtracting 29 from 17 until we obtain a value between -14 and 14

\(\begin{array}{c}a \equiv 17\;(\,\bmod \,\;29)\\ \equiv 17 - 29\;(\,\bmod \,\;29)\\ \equiv - 12\;(\,\bmod \;\,29)\end{array}\)

Since -12 is between -14 and 14:

\(a = - 12\)

Hence the integer a is -12.

Step 4(c): Finding the integer a when \({\bf{a}} =  - {\bf{11}}\;\left( {{\bf{mod}}{\rm{ }}{\bf{21}}} \right)\) and \(90 \le a \le 110\).

\(a \equiv - 11\;(\,\bmod \,\;21)\)

We can a find \(a\) such that \(90 \le a \le 110\)by consecutively subtracting 21 from -11 until we obtain a value between 90 and 110

\(\begin{array}{c}a \equiv - 11\;(\,\bmod \;\,21)\\ \equiv - 11 + 21\;(\,\bmod \;\,21)\\ \equiv 10\;(\,\bmod \,\;21)\\ \equiv 10 + 21\;(\,\bmod \;\,21)\end{array}\)

\(\begin{array}{c} \equiv 31\;(\,\bmod \,\;21)\\ \equiv 31 + 21\;(\,\bmod \;\,21)\\ \equiv 52\;(\,\bmod \,\;21)\\ \equiv 52 + 21\;(\,\bmod \,\;21)\end{array}\)

\(\begin{array}{c} \equiv 73\;(\,\bmod \,\;21)\\ \equiv 73 + 21\;(\,\bmod \,\;21)\\ \equiv 94\;(\,\bmod \;\,21)\end{array}\)

Since 94 is between 90 and 110:

\(a = 94\)

Hence the integer a is 94.