Converting Positive Integers - SS1 Mathematics Lesson Note

Given a positive integer \(n\) to be converted to its equivalent \(R\) in \(mod\ m\), thus,

\(\frac{n}{m} = q\ \ \ remainder\ \ \mathbf{R}\),

Thus, \(n = \mathbf{R}\ (mod\ m)\)

\(n = q.m + R\), where \(q\) is any positive whole number whose product with \(m\) is as close as possible to \(n\) without exceeding n. This is a statement to test if the congruence between \(R\ and\ n\) is \(\mathbf{true}\).

Example: Convert 51 to its equivalent in \(mod\ 10\)

Solution

\(n = 51\ and\ m = 10\)

\[\frac{n}{m} = \ \frac{51}{10} = 5\ remainder\ \mathbf{1}\]

Thus, \(51\ \equiv \mathbf{1\ }(mod\ 10)\) meaning \(51\) is equivalent to \(\mathbf{1}\) in \(mod\ 10\).

Checking with \(n = q.m + R\)

\(51 = 5.10 + 1\)

\(51 = 50 + 1\)

\(51 = 51\), which is true, so we are correct.

 

Example: Find \(x\) in the equation, \(3\ \equiv x\ (mod\ 4)\) (In essence, find the equivalent of ordinary \(3\ \)in\(\ mod\ 4\))

Solution

\(n = 3\) and \(\ m = 4\)

\[\frac{3}{4} = 0\ remainder\ \mathbf{3}\]

Thus, \(3 \equiv \mathbf{3}(mod\ 4)\) and \(x = \mathbf{3}\)