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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