Question

find the H.C.F. of the following pairs of numbers and also express it as a linear combination of them
(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 ​

Answer 1

Step-by-step explanation:

(i) 32 and 54

Solution:

Now, apply Euclid’s Division Lemma on 54 and 32

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 remainder 2

10 = 2 x 5 + 0

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

(ii) 18 and 24

Solution:

Now, apply Euclid’s Division Lemma on 24 and 18

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) 70 and 30

Solution:

Now, apply Euclid’s Division lemma on 70 and 30

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 is 10

(iv) 56 and 88

Solution:

Now, apply Euclid’s Division lemma on 56 and 88

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 is 8

(v) 475 and 495

Solution:

By applying Euclid’s Division lemma on 495 and 475 we get,

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 is 5

(vi) 75 and 243

Solution:

By applying Euclid’s Division lemma on 243 and 75

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 is 3

(vii) 240 and 6552

Solution:

By applying Euclid’s Division lemma on 6552 and 240 we get,

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 is 24

(viii) 155 and 1385

Solution:

By applying Euclid’s Division lemma on 1385 and 155 we get,

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 is 5

(ix) 100 and 190

Solution:

By applying Euclid’s division lemma on 190 and 100 we get,

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 is 10

(x) 105 and 120

Solution:

By applying Euclid’s division lemma on 120 and 105 we get,

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 is 15