Mathematics

 Example:  Find a remainder when $1 !~ + 2  ! ~+ 3  !~ + \cdots ~+ 100  !$ is divided by $12$  Answer:   One can observe that $4! \equiv 24 \equiv 0 ~(mod ~12)$; thus, for $k \geq 4,~ k !  \equiv 4! \cdot 5 \cdot 6 ~\cdots ~k \equiv 0 \cdot  5 \cdot 6 ~\cdots~ k \equiv 0 ~(mod ~12)$. Therefore $1!~ + 2!~ + 3!~ + 4! ~+ \cdots ~+ 100! \equiv 1!~ + 2!~ + 3!~ + ~0 +~ \cdots + ~ 0 \equiv 9 ~(mod ~12) $ i.e. $9 $ a remainder when $1 ! + 2  ! + 3  ! + \cdots + 100  !$ divided by 12.

 Example:  Show that $41$ divide $2^{20} - 1$.   OR   Show that $1$ is remainder when $2^{20}$ is divided by $41$.  OR   Show that $2^{20} \equiv 1 ~(mod ~41)$. Answer:  We have to show that $41$ divide $2^{20} - 1$. Clearly we have $2 \equiv 2 ~(mod ~41) $. By property $6.$ of Theorem , we have  $2^{2} \equiv 4 ~(mod ~41)$. On the same line we have $2^{3} \equiv 8 ~(mod ~41)$. Similarly,  $2^{4} \equiv 16 ~(mod ~41)$ and $2^{5} \equiv 32 ~(mod ~41)$. But $32 \equiv -9 ~(mod ~41)$. Therefore $2^{5} \equiv -9 ~(mod ~41)$. Hence $2^{5} \cdot 2^{5} \equiv (-9) \cdot (-9) ~(mod ~41)$, i.e. $2^{10} \equiv 81 ~(mod ~41)$. But $81 \equiv -1 ~(mod ~41) $, so $2^{10} \equiv -1 ~(mod ~41)$. Again, we have  $2^{10} \cdot 2^{10} \equiv (-1) \cdot (-1) ~(mod ~41)$, i.e. $2^{20} \equiv 1 ~(mod ~41)$. Hence $41$ divide $2^{20} - 1$.

Elementary Properties of Congruences:  Let $n > 1$ be fixed and $a,~ b,~ c,~ d$ be arbitrary integers. Then the following properties hold:  1) $a  \equiv a ~(mod ~n)$. 2)  If $a \equiv b ~(mod ~n)$, then $b \equiv a ~(mod ~n)$.  3) If $a \equiv b ~(mod ~n)$ and $b \equiv c ~(mod ~n)$, then  $a \equiv c ~(mod ~n)$.  4) If $a \equiv b ~(mod ~n)$ and $c \equiv d ~(mod ~n)$, then  $a + c \equiv b + d ~(mod ~n)$ and $a \cdot c \equiv b \cdot d ~(mod ~n)$.  5) If $a \equiv b~ (mod ~n)$, then $a + c \equiv b + c ~(mod ~n)$ and $a \cdot c \equiv b \cdot c ~(mod ~n)$.  6) If $a \equiv b~ (mod ~n)$, then $a^k \equiv  b^k ~(mod ~n)$ for any positive integer $k$. 7)   If $a \equiv b~ (mod ~n)$ and $d> 0$ such that $ d \mid n$, then $a \equiv  b ~(mod ~d)$. Proof:  $1.$ For every integer $a$, we have $a - a = 0$ and $n ~|~ (a -a = 0)$. Therefore by definition of congruences $a  \equiv ...

Theorem:   For given any integers $a $ and $b$, $a \equiv b ~(mod ~n) $ if and only if $a$ and $b$ leave the same nonnegative remainder when  divided by $n$. Proof: Firstly, we suppose that  $a \equiv b ~(mod ~n)$ and we will show that $a$ and $b$ leave the same nonnegative remainder when divided by $n$. Since  $a \equiv b ~(mod ~n)$, we have $a = b + k \cdot n$ for some integer $k$. When we divide $b$ by $n$,  $ b$ leaves a certain remainder say $r$; that is,  $b = q \cdot n + r$, where $0 \leq r < n$. We put $b = q \cdot n + r$ in $a = b + k \cdot n$.  Therefore, $a =  b + k \cdot n = (q \cdot n + r) + k \cdot n = (q + k) \cdot n + r$ that is  $a$ has the same remainder as $b$, when we divide $a$ by $n$.  Conversely, suppose that $a$ and $b$ leave the same nonnegative remainder when divided by $n$. So  we can write $a = q_{1} \cdot n + r$ and $b = q_2 \cdot n + r$, with the  same remainder $0 \leq r < n$. Then $a...

  Definition:  The set of $n$ integers $0, ~1, ~2, ... ,~ n - 1$ is called the set of least nonnegative residues  modulo $n$. In general, a collection of $n$ integers $a_{1},~ a_2, \cdots ,~ a_n$ is said to form a complete set of residues or a complete system of residues modulo $n$ if every integer is congruent modulo $n$ to one and only one of the $a_k$ where $1\leq k \leq n$. We can observe that The set of $5$ integers $0, ~1, ~2, ~3,~ 4$ is  the set of least nonnegative residues modulo $5$ and a collection of $5$ integers $10, ~21, ~37, ~43, ~59$  form a complete set of residues (or  a complete system of residues ) modulo $5$. Observe that any $n$ integers form a complete set of residues modulo $n$ if and only if no two of the integers are congruent modulo $n$. Remark:  Note that for given integer $n > 0$, negative integer can  form a complete set of residues (or  a complete system of residues ) modulo $n$. For example $59 \e...

 Definition: Let $n$ be a fixed positive integer. Two integers $a$ and $b$ are said to be congruent modulo $n$}  denoted by $a \equiv b ~(mod ~n)$ if $n$ divides the difference $a - b$; that is, provided that $a - b = k \cdot n$ for some integer $k$.  Now we give an equivalent  definition of congruent modulo $n$   Definition:  Let $n$ be a fixed positive integer. Two integers $a$ and $b$ are said to be \congruent modulo $n$,  denoted by $a  \equiv  b ~(mod~ n)$  if $b$ is a remainder when $a$ is  divided by $n$.  Let $n = 5$, we can observe that $10  \equiv  0 ~(mod~ 5)$ Since $5$ divides $10-0= 10$. Similarly,  we can see that $21  \equiv  1 ~(mod~ 5)$ Since $5$ divides $21-1= 20$, $37  \equiv  2 ~(mod~ 5)$ Since $5$ divides $37-2= 35$, $43  \equiv  3 ~(mod~ 5)$ Since $5$ divides $43-3= 40$, $59  \equiv  4 ~(mod~ 5)$ Since $5$ divides $59-4= 55$. When $ n \nmi...