If HCF of $3, 6$ is ' $a$ ', HCF of $7, 17$ is ' $b$ ' and HCF of $12, 16$ is ' $c$ ' then arrange $a, b, c$ in ascending order.

a<b<c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

b<c<a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

b<a<c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

c<a<b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Prime factorization of $3 = 1 \times 3$

Prime factorization of $6 = 2 \times 3$

$∴$ HCF of $3, 6 = a = 3$ .

Prime factorization of $7 = 1 \times 7$

Prime factorization of $17 = 1 \times 17$ .

$∴$ HCF of $7, 17 = b = 1$ .

Prime factorization of $12 = 2 \times 2 \times 3$

Prime factorization of $16 = 2 \times 2 \times 2 \times 2$

$∴$ HCF of $12, 16 = c = 4$
Since, $1 < 3 < 4$
Therefore, Ascending order of $a, b, c$ is $b < a < c$
Hence, the correct option is C ( $b < a < c$ ).

Suggest Corrections

Similar questions

If c is the HCF of 'a' and 'b', then find the HCF of a/c and b/c .

x^{6} y^{4} z^{2}

x^{6} y^{4} z^{6}

x^{7} y^{9} z^{4}

x^{a} y^{b} z^{c}

a - b - c

HCF of $(x^{2} - 9)$ and $(x + 3)^{2}$ is 'a'. HCF of $x^{2} + x - 6$ and $x^{2} - 4$ is 'b'. Then $a \times b$ is ____.

r, s, t

a, b, c

L C M

a, b, c

r^{2} s^{4} t^{2}

H C F

a, b, c

r s^{2} t

(a, b, c)

Join BYJU'S Learning Program