Question

The HCF of 248 and 492 is equal to
(a) 2
(b) 3
(c) 4
(d) 5

Answer 1

Hint: To find the HCF of 248 and 492, we will be using a prime factorization method. We have to write each number as a product of its prime factors. Then, we have to list the common factors of both the numbers. The product of all common prime factors is the HCF. We have to use the lower power of each common factor.

Complete step by step solution:
We have to find the HCF of 248 and 492. Let us use the prime factorization method. Firstly, we have to write each number as a product of its prime factors. Let us consider 248. We can write 248 in its prime factorization form as
$248={{2}^{3}}\times 31$
Now let us consider 492.
$492={{2}^{2}}\times 3\times 41$
Now, we have to list the common factors of both the numbers.
Common factors of 248 and 492 are ${{2}^{2}}$ .
The product of all common prime factors is the HCF. We have to use the lower power of each common factor.
Therefore, HCF of 248 and 492 is ${{2}^{2}}=2\times 2=4$

So, the correct answer is “Option c”.

Note: Students can also use the division method to find the HCF. In this method, we have to write the given numbers horizontally by separating it with commas.
$\left| \!{\underline {\,
  248,492 \,}} \right. $
Then, we have to find the smallest prime number which can exactly divide these numbers. We will write this prime number on the left.
$2\left| \!{\underline {\,
  248,492 \,}} \right. $
We will then write the quotients below and repeat these steps till there is no coprime number is left.
$\begin{align}
  & 2\left| \!{\underline {\,
  248,492 \,}} \right. \\
 & 2\left| \!{\underline {\,
  124,246 \,}} \right. \\
 & \text{ }\text{ }\text{ }62,123 \\
\end{align}$
The factors in the left-hand side will be the common prime factors. Then the HCF will be the product of these prime factors.
HCF of 248 and 492 is $2\times 2=4$ .
We can also use a short-cut method.
First we have to divide the largest number by the smallest and write the remainder. Here, we will divide 492 by 248.
$\dfrac{492}{248}\Rightarrow \text{Remainder}=244$
Now, we will divide the divisor of the above stage by the remainder.
$\dfrac{248}{244}\Rightarrow \text{Remainder}=4$
We will repeat the above steps till the remainder becomes 0.
$\dfrac{244}{4}\Rightarrow \text{Remainder}=0$
Then, the divisor of the last stage will be the HCF.
Therefore, HCF of 248 and 492 is 4.