Divisibility, Division, and Remainder Properties

Let $B \neq 0$, and let $A$, $B$, $C$, and $K$ be integers. In a standard Euclidean division setup:

\[
\begin{array}{c|c}
\begin{array}{c} A \\
_\_\quad \vdots \\
\hline
K
\end{array} &
\begin{array}{c}
\quad B \\
\hline \\ C
\end{array}
\end{array} \]

The components are defined as follows:

$A$ is the dividend (bölünen),
$B$ is the divisor (bölen),
$C$ is the quotient (bölüm),
$K$ is the remainder (kalan).

The operation must satisfy these fundamental rules:

$A = B \cdot C + K$ — This identity is called the division algorithm equation.
The remainder is always strictly less than the divisor ($K < B$). It is also non-negative ($K \ge 0$).
If $K = 0$, then $A$ is exactly divisible by $B$ ($B$ divides $A$).
If the remainder is also strictly less than the quotient ($K < C$), the divisor and the quotient can swap places without changing the remainder. That is, if $K < B$ and $K < C$, then:

\[
\begin{array}{c,@{\hspace{2cm}}c,c,c}
\begin{array}{c}
\quad &A \;\;\\
-\quad &\vdots \;\; \\
\hline
&\quad K \;\;
\end{array}
\begin{array}{|c}
\quad B \\
\hline
\quad C
\\
\\
\end{array}
\begin{array}{c}
\quad ve \;\; \;\; \;\; \;\; \;\;
\end{array}
\begin{array}{C}
\quad &A \;\;\\
-\quad &\vdots \;\; \\
\hline
&\quad K \;\;
\end{array}
\begin{array}{|c}
\quad C \\
\hline
\quad B
\\
\\
\end{array}
\end{array} \]

Example:

\[
\begin{array}{c,@{\hspace{2cm}}c,c,c}
\begin{array}{c}
\quad &230 \;\;\\
-\quad &221 \;\; \\
\hline
&\quad 9 \;\;
\end{array}
\begin{array}{|c}
\quad 13 \\
\hline
\quad 17
\\
\\
\end{array}
\begin{array}{c}
\quad ve
\end{array}
\begin{array}{c}
\quad
230=13\cdot17+9
\end{array}
\end{array} \]

\[
\begin{array}{c,@{\hspace{2cm}}c,c,c}
\begin{array}{c}
\quad &230 \;\;\\
-\quad &221 \;\; \\
\hline
&\quad 9 \;\;
\end{array}
\begin{array}{|c}
\quad 17 \\
\hline
\quad 13
\\
\\
\end{array}
\begin{array}{c}
\quad ve
\end{array}
\begin{array}{c}
\quad
230=17\cdot13+9
\end{array}
\end{array} \]

Since the remainder (9) is strictly less than both the divisor (13) and the quotient (17), swapping them yields a perfectly valid alternative division structure with the exact same remainder.

Example:

\[
\begin{array}{c,c}
\begin{array}{c}
\quad 134 \;\;\\
-\quad \vdots \;\; \\
\hline
\quad K \;\;
\end{array}
\begin{array}{|c}
\quad A \\
\hline
\quad 11
\\
\\
\end{array}
\end{array}
\]

In the division operation above, given that $K < 11$, find the sum of $A + K$.

Solution:

Since the remainder $K$ is given to be less than the quotient 11 ($K < 11$), we can swap the positions of the divisor $A$ and the quotient 11. Performing the standard division of 134 by 11:

\[
\begin{array}{c,c}
\begin{array}{c}
\quad 134 \;\;\\
-\quad 132 \;\;\;\; \\
\hline
\;\;\quad 2 \rightarrow K \\
\end{array}
\begin{array}{|c}
\quad 11 \\
\hline
\quad 12 \rightarrow A
\\
\\
\end{array}
\end{array}
\]

Through this swap, we find that $A = 12$ and $K = 2$. Therefore, the requested sum is:
\[ A + K = 12 + 2 = 14 \]

Properties of Remainders:

1. Let $a$ be the remainder when an integer $A$ is divided by $x$, and $b$ be the remainder when an integer $B$ is divided by $x$.

a) The remainder when the sum $A + B$ is divided by $x$ is equal to the remainder of $(a + b) \div x$.

b) The remainder when the product $A \cdot B$ is divided by $x$ is equal to the remainder of $(a \cdot b) \div x$.

Warning: If the calculated values $(a + b)$ or $(a \cdot b)$ turn out to be greater than or equal to $x$, you must divide that resulting sum or product by $x$ again to obtain the final modular remainder.

Examples:

Let’s find the remainder of the sum and the product of the numbers $2357984$ and $1996$ when divided by 9.

First, find the individual remainders using the sum of digits rule for 9:
For $A = 2357984 \rightarrow 2+3+5+7+9+8+4 = 38 \rightarrow 3+8 = 11 \rightarrow 1+1 = 2$. So, $a = 2$.
For $B = 1996 \rightarrow 1+9+9+6 = 25 \rightarrow 2+5 = 7$. So, $b = 7$.

Sum ($A+B$): The sum of their remainders is $a + b = 2 + 7 = 9$. Since 9 is equal to our divisor, we divide it by 9 again: $9 \div 9$ leaves a remainder of 0.

Product ($A \cdot B$): The product of their remainders is $a \cdot b = 2 \cdot 7 = 14$. Dividing 14 by 9 ($14 \div 9$) leaves a final remainder of 5

Uğur groups his candies by 7 and has 5 candies left over. His sister Melek groups her candies by 7 and has 4 candies left over. If Uğur gives all his candies to Melek, find how many candies will be left over when Melek groups the entire batch by 7.

When Melek combines the candies, the total left-over value is calculated by adding the individual remainders: $5 + 4 = 9$. Since 9 is larger than our divisor 7, we evaluate $9 \div 7$, which leaves a remainder of 2. Thus, 2 candies will be left over in the final scenario.

2. If an integer $A$ is exactly divisible by the product $x \cdot y$, it is also separately divisible by both $x$ and $y$. However, the converse is not always true. If $A$ is separately divisible by both $x$ and $y$, it might not be divisible by $x \cdot y$. In such scenarios, $A$ is guaranteed to be divisible by the least common multiple ($\text{LCM}$) of $x$ and $y$. Therefore:

\[
\frac{A}{x \cdot y} \in \mathbb{Z} \implies \frac{A}{x} \in \mathbb{Z} \quad \text{and} \quad \frac{A}{y} \in \mathbb{Z}
\]

But conversely:

\[
\frac{A}{x} \in \mathbb{Z} \quad \text{and} \quad \frac{A}{y} \in \mathbb{Z} \implies \frac{A}{x \cdot y} \in \mathbb{Z} \quad \text{(Not always true)}
\]
\[
\frac{A}{x} \in \mathbb{Z} \quad \text{and} \quad \frac{A}{y} \in \mathbb{Z} \implies \frac{A}{\text{lcm}(x, y)} \in \mathbb{Z} \quad \text{(Always true)}
\]

Example:

The number 320 is divisible by 80. Consequently, it is divisible by all the individual factors of 80.

Conversely, look at this counterexample: the number 320 is exactly divisible by 32 and is also exactly divisible by 20. However, 320 is not divisible by their product $32 \cdot 20 = 640$ (note that 32 and 20 are not coprime). But since $\text{lcm}(32, 20) = 160$, the number 320 is guaranteed to be exactly divisible by 160.