Show that one and only one out of x,(x+1)and (x+2 ) is divisible by 3

1.	Show that one and only one out of x,(x+1)and (x+2 ) is divisible by 3
Answer» On dividing n by 3, let q be the quotient and r be the remainder.Then, {tex}n = 3q + r{/tex}, where {tex}0 \\leq r < 3{/tex}{tex}\\Rightarrow\\;n = 3q + r{/tex} , where r = 0,1 or 2{tex}\\Rightarrow{/tex}{tex}n = 3q \\;or \\;n = (3q + 1) \\;or\\; n = (3q + 2){/tex}.Case I If n = 3q then n is clearly divisible by 3.Case II If\xa0{tex}\\;n = (3q + 1)\\; {/tex} then {tex} (n + 2)= (3q + 1 + 2) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.In this case, {tex}(n + 2){/tex} is divisible by 3.Case III If n = {tex}(3q + 2){/tex} then {tex}(n + 1) = (3q + 2 + 1) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.In this case,{tex} (n + 1){/tex} is divisible by 3.Hence, one and only one out of {tex}n, (n + 1){/tex} and {tex}(n + 2){/tex} is divisible by 3.

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