Question

# 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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## 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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