Question

What is the H.C.F. of \[4 \times 27 \times 3125\], \[8 \times 9 \times 25 \times 7\], \[16 \times 81 \times 5 \times 11 \times 49\]?
A.180
B.360
C.540
D.1260

Answer 1

Hint: Here, we will first express the given numbers as a product of their prime factors. Then, we will take out the common factors from all the three numbers and multiply them to get the required H.C.F.

Complete step-by-step answer:
We have to find the H.C.F. of three numbers which are already given in a form of multiplication of its various factors.
But, as we can see, the factors are not prime factors, i.e. they have at least one factor other than 1 and itself.
Thus, first of all, we will write the three numbers as a product of their prime factors.
So, the first number is: \[4 \times 27 \times 3125\]
Now we will express each number as a product of prime factors.
\[27 = 3 \times 3 \times 3 = {3^3}\]
\[3125 = 5 \times 5 \times 5 \times 5 \times 5 = {5^5}\]
\[4 = 2 \times 2 = {2^2}\]
Thus, the first number can be written as:
\[4 \times 27 \times 3125 = {2^2} \times {3^3} \times {5^5}\]
Similarly, the second number is: \[8 \times 9 \times 25 \times 7\]
Now we will express each number as a product of prime factors.
\[8 = 2 \times 2 \times 2 = {2^3}\]
\[9 = 3 \times 3 = {3^2}\]
\[25 = 5 \times 5 = {5^2}\]
Thus, the second number can be written as:
\[8 \times 9 \times 25 \times 7 = {2^3} \times {3^2} \times {5^2} \times 7\]
Also, the third number is: \[16 \times 81 \times 5 \times 11 \times 49\]
Now we will express each number as a product of prime factors.
\[16 = 2 \times 2 \times 2 \times 2 = {2^4}\]
\[81 = 3 \times 3 \times 3 \times 3 = {3^4}\]
\[49 = 7 \times 7 = {7^2}\]
Thus, the third number can be written as:
\[16 \times 81 \times 5 \times 11 \times 49 = {2^4} \times {3^4} \times 5 \times 11 \times {7^2}\]
Now, we are required to find the H.C.F of these three numbers. A Highest common factor (H.C.F.) is the greatest number that is a factor of two or more numbers. This is also known as the Greatest Common Factor (G.C.F.).
As there three factors are present in all the three numbers and we have to take the lowest power which is common. Therefore,
The H.C.F. of the three numbers \[ = {2^2} \times {3^2} \times 5\]
Applying the exponent on the terms, we get
\[ \Rightarrow \] The H.C.F. of these three numbers \[ = 4 \times 9 \times 5\]
Multiplying the terms, we get
\[ \Rightarrow \] The H.C.F. of these three numbers \[ = 180\]
Therefore, the H.C.F. of \[4 \times 27 \times 3125\], \[8 \times 9 \times 25 \times 7\], \[16 \times 81 \times 5 \times 11 \times 49\] is 180.
Hence, option A is the correct answer.

Note: In this question, in order to find the H.C.F. of the given three numbers, we are required to express the given numbers as a product of their prime factors. Prime factors are those factors which are greater than 1 and have only two factors, i.e.1 and the number itself. Expressing the numbers as a product of prime factors helps us to take out the common factors from the three numbers which when multiplied together, gives us the required H.C.F.