Question

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

Answer 1

ANSWER:

The number is 4.

EXPLANATION:

Find HCF:

76 = 2² x 19

132 = 2² x 3 x 11

160 = 2⁵ x 5

HCF = 2² = 4

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The Greatest possible number that can divide 76, 132 and 160 is 4, which gives a remainder of zero in each case.

Answer 2

Answer:

28

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.