Find the greatest possible number which can divide 76 132 and 160 and leaves remainder same in each case
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Step-by-step explanation:
Suppose the greatest number which can divide 76, 132 and 160 is p and the remainder is r.
Also let q1, q2 and q3 are the quotients of 76, 132 and 160 respectively.
So we have;
pq1+r = 76 ...(i)pq2+r = 132 ...(ii)pq3+r = 160 ...(iii)Now from (i), (ii) and (iii) we get;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 = 2×2×3×7HCF OF 56, 38 and 84 = 2×2×7 = 28
Therefore the required greatest number is 28.
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