Question

Hcf of (2³×3²×5),(2²×3³×5⁵)&(2⁴×3×5³×7×) is

Answer 1

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Answer 2

HCF is 60.

Given:

• Prime factors of three numbers.
• ${2}^{3} \times {3}^{2} \times 5$
• ${2}^{2} \times {3}^{3} \times {5}^{5}$ and
• ${2}^{4} \times 3 \times {5}^{3} \times 7$

To find:

• Find the HCF.

Solution:

Concept/Formula to be used:

HCF: Choose the highest common factor,i.e. the prime numbers which are present in all numbers having the same exponent.

Step 1:

Analyze the prime factors of given numbers.

In all three numbers,

${2}^{3} \times {3}^{2} \times 5 , \: {2}^{2} \times {3}^{3} \times {5}^{5}, \: \: {2}^{4} \times {3} \times {5}^{3} \times 7$

it is clear that, prime numbers 2,3 and 5 are

prime numbers 2,3 and 5 are present in all three numbers, but we can't take 7, because it is not present in all numbers.

Step 2:

Find the HCF of numbers.

Check the maximum common powers of 2,3 and 5.

$HCF= {2}^{2} \times 3 \times 5$

$HCF= 4 \times 15$

$\bf HCF= 60$

Thus,

$\bf HCF({2}^{3} \times {3}^{2} \times 5, \: {2}^{2} \times {3}^{3} \times {5}^{5} \:,{2}^{4} \times {3} \times {5}^{3} \times 7) = 60$

Learn more:

1) FIND THE H.C.F. OF 26 AND 91 BY PRIME FACTORISATION METHOD...

https://brainly.in/question/19449495

2) Hcf of 32 and 144 by prime factorization

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