$1$ and $11$ are the factors of $11$.

The number eleven is a natural number. Mathematically, It is also a whole number because the number $11$ represents a whole quantity.

Sometimes, the number eleven should be expressed as factors of two or more numbers. So, it is essential for everyone to know the factors of $11$ and also to learn how to find the factors of eleven in mathematics.

Arithmetically, $1$ and $11$ are the factors of number $11$. Now, let’s prove mathematically why $1$ and $11$ are only the factors of eleven.

The number $1$ is a first natural number in mathematics. So, let’s divide the number $11$ by $1$ firstly.

$11 \div 1$

$=\,\,$ $\dfrac{11}{1}$

Let’s use the long division method to divide the number $11$ by $1$ and also investigate about the remainder.

$\require{enclose}

\begin{array}{rll}

11 && \hbox{} \\[-3pt]

1 \enclose{longdiv}{11}\kern-.2ex \\[-3pt]

\underline{-~~~11} && \longrightarrow && \hbox{$1 \times 11 = 11$} \\[-3pt]

\phantom{00} 0 && \longrightarrow && \hbox{No Remainder}

\end{array}$

There is no remainder when the number $11$ is divided by $1$. Therefore, it is proved that the number $1$ is a factor of $11$.

Similarly, let’s divide the number eleven by two.

$11 \div 2$

$=\,\,$ $\dfrac{11}{2}$

Let’s see the long division method to divide the number $11$ by $2$ and also know about the remainder.

$\require{enclose}

\begin{array}{rll}

5 && \hbox{} \\[-3pt]

2 \enclose{longdiv}{11}\kern-.2ex \\[-3pt]

\underline{-~~~10} && \longrightarrow && \hbox{$2 \times 5 = 10$} \\[-3pt]

\phantom{00} 1 && \longrightarrow && \hbox{Remainder}

\end{array}$

There is remainder when the number $11$ is divided by $2$, which means the number $2$ does not divide $11$ completely. So, the number $2$ is not a factor of $11$.

Now, let’s use the long division method to divide the number eleven by the natural numbers from $3$ to $10$.

$(1).\,\,$ $11 \div 3$ $\,=\,$ $\dfrac{11}{3}$ $\,\,\longrightarrow\,\,$ $\hbox{Remainder 2}$

$(2).\,\,$ $11 \div 4$ $\,=\,$ $\dfrac{11}{4}$ $\,\,\longrightarrow\,\,$ $\hbox{Remainder 3}$

$(3).\,\,$ $11 \div 5$ $\,=\,$ $\dfrac{11}{5}$ $\,\,\longrightarrow\,\,$ $\hbox{Remainder 1}$

$(4).\,\,$ $11 \div 6$ $\,=\,$ $\dfrac{11}{6}$ $\,\,\longrightarrow\,\,$ $\hbox{Remainder 5}$

$(5).\,\,$ $11 \div 7$ $\,=\,$ $\dfrac{11}{7}$ $\,\,\longrightarrow\,\,$ $\hbox{Remainder 4}$

$(6).\,\,$ $11 \div 8$ $\,=\,$ $\dfrac{11}{8}$ $\,\,\longrightarrow\,\,$ $\hbox{Remainder 3}$

$(7).\,\,$ $11 \div 9$ $\,=\,$ $\dfrac{11}{9}$ $\,\,\longrightarrow\,\,$ $\hbox{Remainder 2}$

$(8).\,\,$ $11 \div 10$ $\,=\,$ $\dfrac{11}{10}$ $\,\,\longrightarrow\,\,$ $\hbox{Remainder 1}$

The numbers $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9$ and $10$ do not divide the number $11$ completely, and each division gives us a remainder. So, the numbers from $3$ to $10$ are not factors of $11$.

Finally, let’s divide the number $11$ by itself. Actually, the number eleven is a factor of itself as per a property of factors.

$11 \div 11$

$=\,\,$ $\dfrac{11}{11}$

Let’s observe the long division process to divide the number $11$ by itself.

$\require{enclose}

\begin{array}{rll}

1 && \hbox{} \\[-3pt]

11 \enclose{longdiv}{11}\kern-.2ex \\[-3pt]

\underline{-~~~11} && \longrightarrow && \hbox{$11 \times 1 = 11$} \\[-3pt]

\phantom{00} 0 && \longrightarrow && \hbox{No Remainder}

\end{array}$

There is no remainder and the factorization has proved that the number $11$ completely divides by itself. So, the number eleven is a factor of itself.

Except $1$ and $11$, the numbers from $2$ to $10$ are failed to divide the number $11$ completely. Therefore, the numbers $1$ and $11$ are the factors of $11$.

The factors of $11$ is expressed in mathematical form as follows.

$F_{11} \,=\, \{1, 11\}$

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