Question

show that one and only one out of n (n+1 ) and (n+2) is divisible by 3,where n is any postive integer​

Answer 1

Answer:

I will explain 2 simple methods to prove this

Method 1:

Simply draw a number line for +ve numbers ,pick any 3 consecutive number , because n, n+1 ,n+2 will be consecutively definitely.

Find the remainder of these , the reminder will be (0,1,2),(1,2,0),(2,1,0) either of them so as you can see exactly one of these number will be divisible by 3. One extra point here is this will also be true even for negative integers.

Method 2:

Method 2 is nothing but explanation of method 1. Just take any integer and if you will divide it by 3 what will be the possible remainder , obviously 0 or 1 or 2.

So I can write any number n as 3*t+x, where n is the number

t is the dividend and x is the remainder .

And as discussed above x can take values only 0, 1,2.

So if we pick n, n+1, n+2 and there remainder will be x, x+1 and x+2.

So now problem is reduced to is exactly 1 of these remainder is divisible by 3.

Think and you will find the possible values of x,x+1,x+2 as (0,1,2) or (1,2,0) or (2,0,1) .So exactly 1 number is divisible by 3.