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Define HCF of two positive integers and find the HCF of the following pairs of numbers.
(i) 32 and 54 (ii) 18 and 24 (iii) 70 and 30
(iv) 56 and 88 (v) 475 and 495 (vi) 75 and 243
(vii) 240 and 6552 (viii) 155 and 1385 (ix) 100 and 190
(x) 105 and 120

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Solution

We need to find H.C.F. of 32 and 54.

By applying division lemma 54 = 32 x 1 + 22

Since remainder ≠ 0, apply division lemma on 32 and remainder 22

32 = 22 x 1 + 10

Since remainder ≠ 0, apply division lemma on 22 and remainder 10

22 = 10 x 2 + 2

Since remainder ≠ 0, apply division lemma on 10 and 2

10 = 2 x 5 + 0

Therefore, H.C.F. of 32 and 54 is 2

(ii) We need to find H.C.F. of 18 and 24.

By applying division lemma

24 = 18 x 1 + 6.

Since remainder ≠ 0, apply division lemma on divisor 18 and remainder 6

18 = 6 x 3 + 0.

Therefore, H.C.F. of 18 and 24 is 6

(iii) We need to find H.C.F. of 70 and 30.

By applying Euclid’s Division lemma

70 = 30 x 2 + 10.

Since remainder ≠ 0, apply division lemma on divisor 30 and remainder 10

30 = 10 x 3 + 0.

Therefore, H.C.F. of 70 and 30 = 10

(iv) We need to find H.C.F. of 56 and 88.

By applying Euclid’s Division lemma

88 = 56 x 1 + 32.

Since remainder ≠ 0, apply division lemma on 56 and remainder 32

56 = 32 x 1 + 24.

Since remainder ≠ 0, apply division lemma on 32 and remainder 24

32 = 24 x 1+ 8.

Since remainder ≠ 0, apply division lemma on 24 and remainder 8

24 = 8 x 3 + 0. Therefore, H.C.F. of 56 and 88 = 8

(v) We need to find H.C.F. of 475 and 495.

By applying Euclid’s Division lemma,

495 = 475 x 1 + 20.

Since remainder ≠ 0, apply division lemma on 475 and remainder 20

475 = 20 x 23 + 15.

Since remainder ≠ 0, apply division lemma on 20 and remainder 15

20 = 15 x 1 + 5.

Since remainder ≠ 0, apply division lemma on 15 and remainder 5

15 = 5 x 3+ 0.

Therefore, H.C.F. of 475 and 495 = 5

(vi) We need to find H.C.F. of 75 and 243.

By applying Euclid’s Division lemma

243 = 75 x 3 + 18.

Since remainder ≠ 0, apply division lemma on 75 and remainder 18

75 = 18 x 4 + 3.

Since remainder ≠ 0, apply division lemma on divisor 18 and remainder 3

18 = 3 x 6+ 0.

Therefore, H.C.F. of 75 and 243 = 3

(vii) We need to find H.C.F. of 240 and 6552.

By applying Euclid’s Division lemma

6552 = 240 x 27 + 72.

Since remainder ≠ 0, apply division lemma on divisor 240 and remainder 72

240 = 72 x 3+ 24.

Since remainder ≠ 0, apply division lemma on divisor 72 and remainder 24

72 = 24 x 3 + 0.

Therefore, H.C.F. of 240 and 6552 = 24

(viii) We need to find H.C.F. of 155 and 1385.

By applying Euclid’s Division lemma

1385 = 155 x 8 + 145.

Since remainder ≠ 0, apply division lemma on divisor 155 and remainder 145.

155 = 145 x 1 + 10.

Since remainder ≠ 0 apply division lemma on divisor 145 and remainder 10

145 = 10 x 14 + 5.

Since remainder ≠ 0, apply division lemma on divisor 10 and remainder 5

10 = 5 x 2 + 0.

Therefore, H.C.F. of 155 and 1385 = 5

(ix) We need to find H.C.F. of 100 and 190.

By applying Euclid’s division lemma

190 = 100 x 1 + 90.

Since remainder ≠ 0, apply division lemma on divisor 100 and remainder 90

100 = 90 x 1 + 10.

Since remainder ≠ 0, apply division lemma on divisor 90 and remainder 10

90 = 10 x 9 + 0.

Therefore, H.C.F. of 100 and 190 = 10

(x) We need to find H.C.F. of 105 and 120.

By applying Euclid’s division lemma

120 = 105 x 1 + 15.

Since remainder ≠ 0, apply division lemma on divisor 105 and remainder 15

105 = 15 x 7 + 0.

Therefore, H.C.F. of 105 and 120 = 15.


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