hcf of 271 and 1238
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Answered by
2
by euclid"s division lemma,
1238=271×4+154
271=154×1+117
154=117×1+37
117=37×3+6
37=6×6+1
6=1×6+0
her the remainder is 0
so the required hcf is 1
1238=271×4+154
271=154×1+117
154=117×1+37
117=37×3+6
37=6×6+1
6=1×6+0
her the remainder is 0
so the required hcf is 1
Answered by
2
by division algorithm
1238=271*4+154
271=154*2+117
154=117*1+37
117=37*3+6
37=6*6+1
6=1*6+0
HCF= 1
1238=271*4+154
271=154*2+117
154=117*1+37
117=37*3+6
37=6*6+1
6=1*6+0
HCF= 1
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