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Using prime factorisation, find the HCF of $144, 180$ and $192$.
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$144 = 2^4 \times 3^2$, $180 = 2^2 \times 3^2 \times 5$, $192 = 2^6 \times 3$
HCF $(144, 180, 192) = 2^2 \times 3 = 12$
HCF $(144, 180, 192) = 2^2 \times 3 = 12$