Question

Find the HCF of 20, 24, 36.

Answer 1

Hint: In this question, we need to find the highest common factor of 20, 24 and 36. For this, we will first find the factor of given numbers. Factors are numbers, which divide the given number. After that, we will find factors, which are common to all three numbers. Then the highest numbers among those common factors will be our required factor.

Complete step-by-step solution
Here we are given numbers 20, 24, and 36.
Now let us find the factors of these three numbers one by one.
For 20,
We will first divide 20 by the smallest prime number which is 2, we get ${\displaystyle \frac{20}{2}}=10$. Now we will divide 10 by the smallest prime number 2, which is 2, we get ${\displaystyle \frac{10}{2}}=5$. 5 is also a prime number. So, prime factorisation of 20 looks like this:
$$\begin{array}{rl}& 2|\underset{―}{20}\\ & 2|\underset{―}{10}\leftarrow \text{20 divided by 2 gives 10}\\ & 5|\underset{―}{5}\leftarrow 1\text{0 divided by 2 gives 5}\\ & |1\leftarrow 5\text{ divided by 5 gives 1}\end{array}$$
So $20=2\times 2\times 5$.
For 24,
Dividing 24 by smallest prime numbers which is 2 we get ${\displaystyle \frac{24}{2}}=12$. Now dividing 12 by 2, we get ${\displaystyle \frac{12}{2}}=6$. Again dividing 6 by 2 we get ${\displaystyle \frac{6}{2}}=3$. 3 is prime number so, prime factorisation of 24 looks like this,
$$\begin{array}{rl}& 2|\underset{―}{24}\\ & 2|\underset{―}{12}\leftarrow \text{24 divided by 2 gives 12}\\ & 2|\underset{―}{6}\leftarrow 1\text{2 divided by 2 gives 6}\\ & 3|\underset{―}{3}\leftarrow 6\text{ divided by 2 gives 3}\\ & |1\leftarrow 3\text{ divided by 3 gives 1}\end{array}$$
So, $24=2\times 2\times 2\times 3$.
For 36,
Dividing 36 by 2 we get ${\displaystyle \frac{36}{2}}=18$. Now dividing 18 by 2 we get 9. Now 9 will not get divided by 2 so the next prime number is 3. Dividing 9 by 3 we get ${\displaystyle \frac{9}{3}}=3$. 3 is prime number. So, prime factorisation of 36 looks like this
$$\begin{array}{rl}& 2|\underset{―}{36}\\ & 2|\underset{―}{18}\leftarrow \text{36 divided by 2 gives 18}\\ & 3|\underset{―}{9}\leftarrow 18\text{ divided by 2 gives 9}\\ & 3|\underset{―}{3}\leftarrow 9\text{ divided by 3 gives 3}\\ & |1\leftarrow 3\text{ divided by 3 gives 1}\end{array}$$
So, $36=2\times 2\times 3\times 3$.
Now from the factorization of 20, 24, and 36, we can see common factors are 2,2.
So the highest common factor of 20, 24, 36 will be $2\times 2=4$.
Hence 4 is the required HCF.
Now the greatest factor among these common factors is 4 so the highest common factor of 20, 24, and 36 is 4.

Note: Students should keep in mind, we need to take common factors from all three numbers. They can make the mistake of taking common factors only from two of the numbers. Students can use the divisibility rule of 2, 3, and 5 to check if a number is divisible by 2, 3, or 5. The divisibility rule of 2 states that the digit at one place should be 0, 2, 4, 6, 8. The divisibility rule of 3 states that the sum of digits of a number should be also divisible by 3. The divisibility rule of 5 states that digit at one place should be 0 or 5.