Q2/124) Show that in a sequences of 3 numbers one number is always divisible by 3

This can be proved by pigeon hole principle. As there are 3 numbers there are 3 remainders and the remainders can be 0 or 1 or 2 and no 2 remainder can be same if they where same then difference is divsible by 3 but difference cannot be be >2 so this is not possible.

So one of the remainders has to be zero.

Hence proved.

However this can be proved as below

Add 3 to all the terms starting from 1^st number and keep adding. Then we get all the terms to the right. Subtract 3 from 3^rd term and keep subtracting. We get all the numbers and none of them is divisible by 3. which is a contradiction. So one of the numbers has to be divisible by 3.