HCF of 3 and 6 is 'a', HCF of 7 and 17 is 'b' and HCF of 12 and 16 is 'c'. Arrange a, b and c in ascending order.

a<b<c

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b<c<a

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b<a<c

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c<a<b

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Solution

The correct option is C
b<a<c

Factorization

3 = $1 \times 3$ and 6 = $2 \times 3$ . so HCF is a = 3.

7 = $1 \times 7$ and 17 = $1 \times 17$ . so HCF is b= 1.

12 = $2^{2} \times 3$ and 16 = $2^{4}$ . so HCF is c = 4

Ascending order b<a<c

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16

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