# Find the HCF of the following.

12, 16 and 28.

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**Hint:**HCF stands for Highest Common Factor. It means that if we factorise all the given numbers, the highest factor that they all will have in common will be our HCF. We have been given the numbers 12, 16 and 28. To find their HCF, we will use the Euclid’s Division Lemma. It is used to find the HCF of two positive integers ‘a’ and ‘b’. According to Euclid’s Division Lemma, two positive integers ‘a’ and ‘b’ can be written as $a=bq+r,0\le r < b$ . Using this, we will first find the HCF of any two of the given three numbers and then we will find the HCF of the third number and the HCF of the first two numbers. Then obtained HCF will be our answer.

**Complete step by step answer:**

Now, we have the numbers 12, 16 and 28.

We will take the numbers and 12 and 16 first and find their HCF.

Now, according to Euclid’s Division Lemma, $a=bq+r,0\le r < b$ .

We apply this until we get the remainder ‘r’ as ‘0’ taking ‘b’ and ‘r’ as the next ‘a’ and ‘b’ respectively and the value of ‘b’ for which the value of ‘r’ is 0 will be the required HCF.

Here, $16>12$ , therefore, $a=16$ and $b=12$

Applying the division lemma to 16 and 12 we get:

$16=12\left( 1 \right)+4\text{ }\left( \text{here, q=1} \right)$

Since, $r=4\ne 0$, we will apply the division lemma again for 12 and 4.

Applying Euclid’s division lemma for 12 and 4 we get:

$12=4\left( 3 \right)+0\left( \text{here, q=3} \right)$

Here, $r=0$ and $b=4$ .

Thus, the HCF of 12 and 16 is ‘4’.

Now that we have the HCF of ’12’ and ‘16’ as ‘4’, we can find the HCF of ‘4’ and ‘28’ and it will be our required HCF.

Here, $28 > 4$ , therefore, $a=28$ and $b=4$

Applying Euclid’s Division Lemma for 4 and 28 we get

$28=4\left( 7 \right)+0\left( \text{here, q=7} \right)$

Here, $r=0$ and $b=4$.

Thus, the HCF of ‘4’ and ‘28’ is ‘4’.

**Thus, our required HCF is ‘4’.**

**Note:**This question can also be done by the following method:

We can factorise 12, 16 and 28 separately and then multiply all their prime factors. Hence, we will get our required HCF.

Factorisation of 12 is shown below:

$\begin{align}

& 2\left| \!{\underline {\,

12 \,}} \right. \\

& 2\left| \!{\underline {\,

6 \,}} \right. \\

& 3\left| \!{\underline {\,

3 \,}} \right. \\

& 1\left| \!{\underline {\,

1 \,}} \right. \\

\end{align}$

Thus, 12 can be written as:

$12={{2}^{2}}\times 3$

Factorisation of 16 is shown below:

$\begin{align}

& 2\left| \!{\underline {\,

16 \,}} \right. \\

& 2\left| \!{\underline {\,

8 \,}} \right. \\

& 2\left| \!{\underline {\,

4 \,}} \right. \\

& 2\left| \!{\underline {\,

2 \,}} \right. \\

& 1\left| \!{\underline {\,

1 \,}} \right. \\

\end{align}$

Thus, 16 can be written as:

$16={{2}^{4}}$

Factorisation of 28 is shown below:

$\begin{align}

& 2\left| \!{\underline {\,

28 \,}} \right. \\

& 2\left| \!{\underline {\,

14 \,}} \right. \\

& 7\left| \!{\underline {\,

7 \,}} \right. \\

& 1\left| \!{\underline {\,

1 \,}} \right. \\

\end{align}$

Thus, 28 can be written as:

$28={{2}^{2}}\times 7$

Here, we can see that ${{2}^{2}}$ is a common factor for all the three numbers.

Thus, $HCF={{2}^{2}}=4$