Math, asked by Avni7946, 11 months ago

Find the greatest possible number which can divide 76 132 and 160 and leaves remainder same in each case

Answers

Answered by Avni2348
18

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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