Question

find the greatest possible no. which can divide 76, 132 and 160 in each case​

Answer 1

Answer:

4

Step-by-step explanation:

hcf of 76,132,160=4

Answer 2

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

= 2×2×3×7

HCF OF 56, 38 and 84 = 2×2×7

= 28

Therefore the required greatest number is 28.