HCF of (23 × 32 × 5), (22 × 33 ×52)

Question:

HCF of $\left(2^{3} \times 3^{2} \times 5\right),\left(2^{2} \times 3^{3} \times 5^{2}\right)$ and $\left(2^{4} \times^{3} \times 5^{3} \times 7\right)$ is

(a) 30
(b) 48
(c) 60
(d) 105

Solution:

(c) 60

HCF $=\left(2^{3} \times 3^{2} \times 5,2^{2} \times 3^{3} \times 5^{2}, 2^{4} \times 3 \times 5^{3} \times 7\right)$

HCF = Product of smallest power of each common prime factor in the numbers

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

$=60$

 

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