Find the greatest possible no. which can divide 76,132 and 160 and leaving the same remainder in each case.

Suppose the greatest number which can divide 76, 132 and 160 is p and the remainder is r.
Also let $q_1, q_2 \mathrm{and} q_3$ are the quotients of 76, 132 and 160 respectively.
So we have;
$$\begin{gathered} p{q}_1+r = 76 ...\left(\mathrm{i}\right) \\ p{q}_2+r = 132 ...\left(\mathrm{ii}\right) \\ p{q}_3+r = 160 ...\left(\mathrm{iii}\right) \\ \mathrm{Now} \mathrm{from} \left(\mathrm{i}\right), \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right) \mathrm{we} \mathrm{get}; \\ p\left(q_2-q_1\right) = 132-76 = 56 \\ p\left(q_3-q_2\right) = 160-132 = 28 \\ p\left(q_3-q_1\right) = 160-76 = 84 \end{gathered}$$
Therefore HCF of 56, 28 and 84 is given by;
$$\begin{gathered} 56 = 2\times 2\times 2\times 7 \\ 28 = 2\times 2\times 7 \\ 84 = 2\times 2\times 3\times 7 \\ \mathrm{HCF} \mathrm{OF} 56, 38 \mathrm{and} 84 = 2\times 2\times 7 = 28 \end{gathered}$$
Therefore the required greatest number is 28.