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Using euclid algorithm to find the HCF of 595 and 252 and then express it in the form of 595m + 252n |
| Answer» Given integers,595 and 252Applying Euclid division algorithm to 595 and 252 we get,595=252×2+91.........................................1Now applying ,Euclid division algorithm to 252 and 91 we get,252=91×2+70.....….………...........................2Now applying Euclid division algorithm to 91 and 70,we get91=70×1+21…...............................................3Now applying Euclid division algorithm to 70 and 21 we get,70=21×3+7...................................................4Now applying Euclid division algorithm to 21 and 7 we get,21=7×3+0......................................................5The remainder at this stage is zero.Hence HCF of 595 and 252 is 7.eq1,eq2,eq3, and eq4 can be written as,595-(252×2)=91..........................................6252-(91×2)=70.............................................791-(70×1)=21...................................................870-(21×3)=7.....................................................9Now eq9 we have,(as we express HCF in the form of a equation that is the reason we start from eq9)7=70-(21×3)By putting eq8 in above equation we have,7=70-(91-70×1)37=70-91×3+70×37=70×4-91×3By putting eq7 in above equation we have,7=(252-91×2)4-91×37=252×4-91×8-91×37=252×4-91×11By putting eq6 in above equation we have,7=252×4-(595-252×2)117=252×4-595×11+252×227=252×26-595×117=595(-11)+252(26)7=595m+252n, where m=-11 and n=26.Hence HCF of 595 and 252 is in the form of 595m+252n,where m=-11 and n=26. | |