Question 65

The HCF of 56, 140 and 168 is:

Solution

Expressing the number as product of prime factors,

56 - 2$$^{3}$$ x 7

140 - 2$$^{2}$$ x 5 x 7

168 - 2$$^{3}$$ x 3 x 7

HCF is obtained by multiplying each common prime factor using the least power power of each prime factor.

Here common prime factors are 2, 7

Least power of 2 here is 2$$^{2}$$

Least power of 7 here is 7$$^{1}$$

HCF = 4x 7 = 28

Hence, option A is the correct answer.

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