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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
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.
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