Prove that one and only one out of n,

Solution:

Let a = n, b = n + 2 and c =n + 4

Order triplet is (a, b, c) = (n, n + 2, n + 4)                                                                      … (i)

Where, n is any positive integer i.e.,n = 1,2, 3,…

At n = 1;   (a, b, c) = (1,1 + 2,1 + 4) = (1, 3, 5)

At n = 2;   (a, b, c) = (2,2 + 2,2 + 4)= (2, 4, 6)

At n = 3;  (a, b,c) =  (3, 3 + 2,3 + 4) = (3,5,7)

At n =4;   (a,b, c) =(4, 4 + 2, 4 +4)  = (4, 6, 8)

At n = 5;   (a, b,c) = (5, 5 + 2, 5 +4)= (5,7,9)

At n = 6;   (a,b, c) =   (6, 6 + 2, 6 + 4)= (6,8,10)        ‘

At n = 7;   (a, b,c) =  (7,7 + 2,7 + 4)= (7, 9,11)

At n = 8;   (a, b,c) =  (8,8+ 2, 8+ 4)= (8,10,12)

We observe that each triplet consist of one and only one number which is multiple of 3 i.e., divisible by 3.

Hence, one and only one out of n,(n + 2) and (n + 4) is divisible by 3, where, n is any positive integer.

Alternate Method

On dividing ‘n’ by 3, letg be the quotient and r be the remainder.

Then, n =,3 q  + r, where, 0< r< 3

$\Rightarrow \quad n=3 q+r$, where, $r=0,1,2$

$\Rightarrow \quad n=3 q$ or $n=3 q+1$ or $n=3 q+2$

Case I If $n=3 a$, then $n$ is only divisible by 3 .

but $n+2$ and $n+4$ are not divisible by 3

Case II If $n=3 q+1$, then $(n+2)=3 q+3=3(q+1)$, which is only divisible by 3 ,

but $n$ and $n+4$ are not divisible by 3 .

So, in this case, $(n+2)$ is divisible by 3 .

Case III When $n=3 q+2$, then $(n+4)=3 q+6=3(q+2)$, which is only divisible by 3 ,

but $n$ and $(n+2)$ are not divisible by 3 .

So, in this case, $(n+4)$ is divisible by 3 .

Hence, one and only one out of n, (n + 2) and (n + 4) is divisible by 3.