Show that exactly one of the number n, n+2 ,or n+4 is divisible by 3
Let n =3kthen n + 2 = 3k + 2and n + 4 = 3k + 4Case 1: When n=3k ,n is divisible by 3 …………(1)n + 2 = 3k + 2or, n + 2 is not divisible by 3n + 4 = 3k + 4= 3(k + 1) + 1or, n + 4 is not divisible by 3Case 2:When n=3k+1, n is not divisible by 3\xa0n + 2 = (3k + 1) + 2=3k + 3 = 3(k + 1){tex} \\Rightarrow{/tex}\xa0n+ 2 is clearly divisible by 3……………………..(2)n + 4 = (3k + 1) + 4= 3k + 5= 3(k + 1) + 2{tex}\\Rightarrow{/tex}\xa0n + 4 is not divisible by 3Case 3:When n=3k+2,n is not divisible by 3\xa0n + 2 = (3k + 2) + 2= 3k + 4(n + 2) is not divisible by 3x + 4 = 3k + 6 = 3(k + 2){tex}\\Rightarrow{/tex}\xa0n + 4 is divisible by 3……………………(3)Hence, from (1),(2) and (3) it is clear that\xa0exactly one of the numbers n, n + 2, n + 4, is divisible by 3.
We applied Euclid Division algorithm on n and 3. a = bq +r on putting a = n and b = 3 n = 3q +r , 0