find h for of 210 and 55 by euclids division
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HCF of 210 and 55 is 5..........
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by euclid"s division lemma,
210=55×3+45
55=45×1+10
45=10×4+5
10=5×2+0
here the remainder is zero
so the 5 is required hcf
210=55×3+45
55=45×1+10
45=10×4+5
10=5×2+0
here the remainder is zero
so the 5 is required hcf
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