HCF of (2^3 × 3^2 × 5), (2^2 × 3^3 × 5^2 ) and (2^4 × 3 × 5^3 × 7)

By BYJU'S Exam Prep

Updated on: September 25th, 2023

To find: HCF of (23×32×5),(22×33×52), (24×3×53×7)

HCF of these numbers is the product of each common prime factor’s smallest power. So, first we will find the common prime factors that are 2, 3, and 5. These prime numbers are common in all set.

Now, we have to find the smallest power of 2, 3, and 5 in (23×32×5),(22×33×52), (24×3×53×7). The smallest power of 2 in the given set is 22. Similarly the smallest power of 3 and 5 are 3 and 5 respectively.

After getting the smallest power of each common prime factor’s we have multiply them in order to obtain the HCF of (23×32×5),(22×33×52), (24×3×53×7).

HCF = product of each common prime factor’s smallest power.

So, HCF of (23×32×5),(22×33×52), (24×3×53×7) = 2x 3 x 5 = 60.

What is HCF of (23×32×5), (22×33×52), (24×3×53×7)?

As discussed above, the HCF for (23×32×5), (22×33×52), (24×3×53×7) is 60. Candidates should be aware of the method to find the HCF of such type of numbers. In these type of questions, candidates need to find the prime number first and after that their smallest power. If a candidate has this concept then he or she can solve this question in a minute.

An expression known as power illustrates repeated multiplication of the same number or factor. The exponent’s value depends on how many times the base has been multiplied by itself.

For example – 8 x 8 x 8 can be written as 83 because 8 is multiplied by itself 3 times. In this case, 3 is the exponent or power that indicates how many times the number 8 has been multiplied by itself, and 8 is the ase that denotes the number being multiplied.

Summary:

HCF of (23×32×5), (22×33×52), (24×3×53×7)

The HCF of (23×32×5), (22×33×52), (24×3×53×7) is 60. In these types of questions, candidates must first determine their prime number and then their smallest power. After determining the prime numbers with the smallest power, multiply them to obtain the correct answer.

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