# Find the HCF of the following numbers :(a) 18, 48(b) 30, 42 (c) 18, 60 (d) 27, 63(e) 36, 84 (f) 34, 102 (g) 70, 105, 175(h) 91, 112, 49 (i) 18, 54, 81 (j) 12, 45, 75

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To do :

We have to find the HCF of the given pairs of numbers.

Solution :

(a) 18,48

Prime factorisation of 18 and 48 is,

$18 = 2\times 3\times 3$

$48 = 2\times 2\times 2\times 2\times 3$

Therefore,

HCF of 18 and 48 $= 2\times 3 = 6$.

Therefore, the HCF of 18 and 48 is 6.

(b) 30,42

Prime factorisation of 30 and 42 is,

$30 = 2\times 3\times 5$

$42 = 2\times 3\times 7$

Therefore,

HCF of 30 and 42 $= 2\times 3 = 6$.

Therefore, the HCF of 30 and 42 is 6.

(c) 18,60

Prime factorisation of 18 and 60 is,

$18 = 2\times 3\times 3$

$60 = 2\times 2\times 3\times5$

Therefore,

HCF of 18 and 60 $= 2\times 3 = 6$.

Therefore, the HCF of 18 and 60 is 6.

(d) 27,63

Prime factorisation of 27 and 63 is,

$27 = 3\times 3\times 3$

$63 = 3\times 3\times 7$

Therefore,

HCF of 27 and 63 $= 3\times 3 = 9$.

Therefore, the HCF of 27 and 63 is 9.

(e) 36,84

Prime factorisation of 36 and 84 is,

$36= 2\times 2\times 3\times3$

$84 = 2\times 2\times 3\times7$

Therefore,

HCF of 36 and 84 $= 2\times2\times3 = 12$.

Therefore, the HCF of 36 and 84 is 12.

(f) 34,102

Prime factorisation of 34 and 102 is,

$34 = 2\times 17$

$102 = 2\times 3\times 17$

Therefore,

HCF of 34 and 102 $= 2\times 17 = 34$.

Therefore, the HCF of 34 and 102 is 34.

(g) 70, 105, 175

Prime factorisation of 70, 105 and 175 is,

$70 = 2\times 5\times7$

$105 = 3\times 5\times 7$

$175 = 5\times 5\times 7$

Therefore,

HCF of 70, 105 and 175 $= 5\times 7 = 35$.

Therefore, the HCF of 70, 105 and 175 is 35.

(h) 91, 112, 49

Prime factorisation of 91, 112 and 49 is,

$91 = 7\times 13$

$112 = 2\times 2\times 2\times 2\times 7$

$49 = 7\times 7$

Therefore,

HCF of 91, 112 and 49 $=7$.

Therefore, the HCF of 91, 112 and 49 is 7.

(i) 18, 54, 81

Prime factorisation of 18, 54 and 81 is,

$18 = 2\times 3\times3$

$54 = 2\times 3\times 3\times3$

$81 = 3\times 3\times 3\times3$

Therefore,

HCF of 18, 54 and 81 $= 3\times 3 = 9$.

Therefore, the HCF of 18, 54 and 81 is 9.

(j) 12, 45, 75

Prime factorisation of 12, 45 and 75 is,

$12 = 2\times 2\times3$

$45 = 3\times 3\times 5$

$75 = 3\times 5\times 5$

Therefore,

HCF of 12, 45 and 75 $= 9$.

Therefore, the HCF of 12, 45 and 75 is 3.

Updated on 10-Oct-2022 13:30:51