using Euclid's division algorithm find whether the pair of number 847,2160 are compromise or not
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Answered by
5
2160=847x2+466
847=466x1+381
466=381x1+85
381=85x4+41
85=41x2+3
41=3x13+2
3=2x1+1
2=1x2+0
therefore HCF is 1
847=466x1+381
466=381x1+85
381=85x4+41
85=41x2+3
41=3x13+2
3=2x1+1
2=1x2+0
therefore HCF is 1
Answered by
3
2160=847×2+466
847=466 ×1+381
466=381×1+85
381= 85×4+41
85= 41×2+3
41=3×13+2
3 = 2×1 +1
2=1×2 +0
1 is hcf.
847=466 ×1+381
466=381×1+85
381= 85×4+41
85= 41×2+3
41=3×13+2
3 = 2×1 +1
2=1×2 +0
1 is hcf.
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