Find the H.C.F of : 5 kg 4Hg,7 kg 2Hg.
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3
Answer:
3kg 7hg 8dag 1g=3781
9kg 1hg 5dag 4g=9154
The greatest weight required is G.C.F. of 3781 and 9154
Using Euclid algorithm
9154=3781×2+1592
3781=1592×2+597
1592=597×2+398
597=398×1+199
398=199×2+0
Hence greatest weight =1hg 9dag 9g
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0
Answer:
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