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R 1135 str 225 () 196 sht 3820(ii) 867 3255) 135 225 |
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Answer» Answer :(i) 135 and 225Since 225 > 135, we apply the division lemma to 225 and 135 to obtain225 = 135 × 1 + 90Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain135 = 90 × 1 + 45We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain90 = 2 × 45 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 45,Therefore, the HCF of 135 and 225 is 45. (ii) 196 and 38220Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain38220 = 196 × 195 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 196,Therefore, HCF of 196 and 38220 is 196. (iii) 867 and 255Since 867 > 255, we apply the division lemma to 867 and 255 to obtain867 = 255 × 3 + 102Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain 255 = 102 × 2 + 51We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain102 = 51 × 2 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51. |
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