The aim of this question is to **introduce** the concept of **congruency** of an integer with another integer **under some modulo**.

Whenever we **divide one integer over another**, we have two results, namely a **quotient** and a **remainder**. The **quotient** is the part of the result that defines the **perfect division** while the existence of the **remainder** signifies that the **division was not perfect**.

Let’s say we have t**hree integers a, b, and c**. Now we say that **a is congruent to b modulo c** if $ a \ – \ b $ is **perfectly divisible** by $ c $.

## Expert Answer

Given that we need to find **all integers** (say $ x $) that are **congruent to 4 modulo 12**. In simpler words, we need to find the **first five values** of $ x \ – \ 4 $ that are **perfectly divisible** by $ 12 $.

To solve this question, we can take help from the **integral multiples** of $ 12 $ as listed below:

\[ \text{ Integral multiples of } 12 \ = \ \{ 0, \ 12, \ 24, \ 36, \ 48, \ 60, \ … \ … \ … \ \} \]

To find the first five integer values that are congruent to 4 modulo 12, we simply need to **solve the following equations**:

\[ \begin{array}{ c } \text{ Integers congruent } \\ \text{ to } 4 \text{ modulo } 12 \end{array} \ = \ \left \{ \begin{array}{ c c c } x \ – \ 4 \ = \ 0 & \Rightarrow & x \ = \ 0 \ + \ 4 & \Rightarrow & x \ = \ 4 \\ x \ – \ 4 \ = \ 12 & \Rightarrow & x \ = \ 12 \ + \ 4 & \Rightarrow & x \ = \ 16 \\ x \ – \ 4 \ = \ 24 & \Rightarrow & x \ = \ 24 \ + \ 4 & \Rightarrow & x \ = \ 28 \\ x \ – \ 4 \ = \ 36 & \Rightarrow & x \ = \ 36 \ + \ 4 & \Rightarrow & x \ = \ 40 \\ x \ – \ 4 \ = \ 48 & \Rightarrow & x \ = \ 48 \ + \ 4 & \Rightarrow & x \ = \ 52 \end{array} \right. \]

\[ \text{ Integers congruent to } 4 \text{ modulo } 12 \ = \ \{ 4, \ 16, \ 28, \ 40, \ 52 \ \} \]

## Numerical Results

\[ \text{ Integers congruent to } 4 \text{ modulo } 12 \ = \ \{ 4, \ 16, \ 28, \ 40, \ 52 \ \} \]

## Example

List down the **first six integers** such that they are **congruent to 5 modulo 15**.

**Here:**

\[ \text{ Integral multiples of } 15 \ = \ \{ 0, \ 15, \ 30, \ 45, \ 60, \ 75, \ … \ … \ … \ \} \]

**So:**

\[ \begin{array}{ c } \text{ Integers congruent } \\ \text{ to } 5 \text{ modulo } 15 \end{array} \ = \ \left \{ \begin{array}{ c c c } x \ – \ 5 \ = \ 0 & \Rightarrow & x \ = \ 0 \ + \ 5 & \Rightarrow & x \ = \ 5 \\ x \ – \ 5 \ = \ 15 & \Rightarrow & x \ = \ 15 \ + \ 5 & \Rightarrow & x \ = \ 20 \\ x \ – \ 5 \ = \ 30 & \Rightarrow & x \ = \ 30 \ + \ 5 & \Rightarrow & x \ = \ 35 \\ x \ – \ 5 \ = \ 45 & \Rightarrow & x \ = \ 45 \ + \ 5 & \Rightarrow & x \ = \ 50 \\ x \ – \ 5 \ = \ 60 & \Rightarrow & x \ = \ 60 \ + \ 5 & \Rightarrow & x \ = \ 65 \end{array} \right. \]

\[ \text{ Integers congruent to } 5 \text{ modulo } 15 \ = \ \{ 5, \ 20, \ 35, \ 50, \ 65 \ \} \]