Question

Define HOE 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

Answer 1

By applying Euclid’s division lemma
(i) 54 = 32 × 1 + 22
Since remainder ≠ 0, apply division lemma on division of 32 and remainder 22.
32 = 22 × 1 + 10
Since remainder ≠ 0, apply division lemma on division of 22 and remainder 10.
22 = 10 × 2 + 2
Since remainder ≠ 0, apply division lemma on division of 10 and remainder 2.
10 = 2 × 5 [remainder 0]
(ii) By applying division lemma
24 = 18 × 1 + 6
Since remainder = 6, apply division lemma on divisor of 18 and remainder 6.
18 = 6 × 3 + 0
∴ Hence, HCF of 18 and 24 = 6
(iii) By applying Euclid’s division lemma
70 = 30 × 2 + 10
Since remainder ≠ 0, apply division lemma on divisor of 30 and remainder 10.
30 = 10 × 3 + 0
∴ Hence HCF of 70 and 30 is = 10.
(iv) By applying Euclid’s division lemma
88 = 56 × 1 + 32
Since remainder ≠ 0, apply division lemma on divisor of 56 and remainder 32.
56 = 32 × 1 + 24
Since remainder ≠ 0, apply division lemma on divisor of 32 and remainder 24.
32 = 24 × 1 + 8
Since remainder ≠ 0, apply division lemma on divisor of 24 and remainder 8.
24 = 8 × 3 + 0
∴ HCF of 56 and 88 is = 8.
(v) By applying Euclid’s division lemma
495 = 475 × 1 +20
Since remainder ≠ 0, apply division lemma on divisor of 475 and remainder 20.
475 = 20 × 23 + 15
Since remainder ≠ 0, apply division lemma on divisor of 20 and remainder 15.

20 = 15 × 1 + 5
Since remainder ≠ 0, apply division lemma on divisor of 15 and remainder 5.
15 = 5 × 3 + 0
∴ HCF of 475 and 495 is = 5.
(vi) By applying Euclid’s division lemma
243 = 75 × 3 + 18
Since remainder ≠ 0, apply division lemma on divisor of 75 and remainder 18.
75 = 18 × 4 + 3
Since remainder ≠ 0, apply division lemma on divisor of 18 and remainder 3.
18 = 3 × 6 + 0
∴ HCF of 243 and 75 is = 3.
(vii) By applying Euclid’s division lemma
6552 = 240 × 27 + 72
Since remainder ≠ 0, apply division lemma on divisor of 240 and remainder 72.
210 = 72 × 3 + 24
Since remainder ≠ 0, apply division lemma on divisor of 72 and remainder 24.
72 = 24 × 3 + 0
∴ HCF of 6552 and 240 is = 24.
(viii) By applying Euclid’s division lemma
1385 = 155 × 8 + 145
Since remainder ≠ 0, applying division lemma on divisor 155 and remainder 145
155 = 145 × 1 + 10
Since remainder ≠ 0, applying division lemma on divisor 10 and remainder 5
10 = 5 × 2 + 0
∴ Hence HCF of 1385 and 155 = 5.
(ix) By applying Euclid’s division lemma
190 = 100 × 1 + 90
Since remainder ≠ 0, applying division lemma on divisor 100 and remainder 90.
90 = 10 × 9 + 0
∴ HCF of 100 and 190 = 10
(x) By applying Euclid’s division lemma
120 = 105 × 1 + 15
Since remainder ≠ 0, applying division lemma on divisor 105 and remainder 15.
105 = 15 × 7 + 0
∴ HCF of 105 and 120 = 15