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Findnumber whichthegreatestgreatestpossiblepossi960 and 1can divide 176, 1152. andin each ease;remainder w same in near |
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Answer» Suppose the greatest number which can divide 76, 132 and 160 ispand the remainder isr. Also letq1,q2andq3are the quotients of 76, 132 and 160 respectively. So we have; pq1+r=76...(i)pq2+r=132...(ii)pq3+r=160...(iii) Nowfrom(i),(ii)and(iii)weget;p(q2−q1)=132−76=56p(q3−q2)=160−132=28p(q3−q1)=160−76=84 Therefore HCF of 56, 28 and 84 is given by; 56=2×2×2×728=2×2×784=2×2×3×7 HCFOF56,38and84=2×2×7=28Therefore the required greatest number is 28 |
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