show that one and only one out of n, n+2, n+4 is divisible by 3

The remainder that ‘n+4’ leaves on division by 3 will be the same as that left by ‘n+1’. This is because
n+4 = (n+1) + 3.
So the question now becomes ‘Show that among n, n+2 and n+1’ one and only one is divisible by 3 i.e. in every three consecutive natural numbers exactly one is divisible by 3.
Suppose n is divisible by 3 then n+1 and n+2 being 1 and 2 more respectively, will leave remainders 1 and 2 respectively i.e they are not divisible by 3.
Suppose n leaves a remainder of 1 on division by 3 then n+1 and n+2 being 1 and 2 more respectively, will leave remainders 2 and 3(or effectively zero) respectively i.e only n +2 is divisible by 3.
Similarly the third case where n leaves a remainder of 2 on division by 3 and hence where only n+1 is divisible by 3

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 We know that any positive integer is of the form 3q,3q+1,3q+2 for some integer q 

    CASE 1

When n =3q , is divisible by 3 

n+2= 3q+2 , is not divisible by 3 as it leaves a remainder 2

n+4 = 3q+4, 3q + 3 +1, 3(q+1)+1, not divisible by 3 as it leaves a remainder 1

Therefore n is divisible by 3 and n+2 and  n+4 are not divisible by 3.

CASE 2

n= 3q +1 noy divisible by 3 because it leaves a remainder 1

n+2= 3q+1+2, 3q+3, 3(q+1) , divisible by 3 

n + 4 = 3q+1+4 , 3q+5 , 3q +3+2,3(q+1)+2 ,not divisible by 3 because it leavesa remainder 2

Therefore n +2 is divisible by 3 and n and n+4 are not divisible by 3

CASE 3

n=3q+2 ,not divisible by 3 because it leaves a remainder 2

n+2= 3q+2+2 , 3q+4 , 3q + 3+1, 3 (q+1)+1 ,not divisible by 3 as it leaves a remainder 1 

n+4=3q+2+4, 3q+6 ,3(q+2),divisible by 3 

Therefore n+4 is divisible by 3 but n and n +2 are not divisible by 3

Therefor only one out of n , n+2, n+4 is divisible by 3 

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