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Sia ? 6 years, 4 months ago
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 {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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