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Show that the one and only one out of n , n + 2 or n + 4 is divisible by 3, Where n is any positive integer

hi i hope this anser is correct
By euclids divison lemma 
a=bq+r
since it is divided by3 the value of b=3
so the possible remainders are 0 1 2
a=3q or 3q+1 or 3q+2
case 1
a=3q{3q is divisible by 3}
a=3q+2{3q+2 is not divisible by 3}
a=3q+4{3q+4 is not divisible by 3}

case 2
a=3q+1{not divisible by 3}
a=3q+1+2{divisible by 3}
a=3q+1+4{not divisible by 3}

case 3
a=3q+2{not divisible by 3}
a=3q+2+2{not divisible by 3}
a=3q+2+4{divisible by 3}

FROM THIS WE CAN UNDERSTAND THAT IN EACH CASE N N+2 N+4 CAME BE DIVISIBLE BY 3 IN ONLY ONE CASE THAN OTHER CASES

 
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Let n = 3k, 3k + 1 or 3k + 2.
(i) When n = 3k:
n is divisible by 3.
n + 2 = 3k + 2  n + 2 is not divisible by 3.
n + 4 = 3k + 4 = 3(k + 1) + 1  n + 4 is not divisible by 3.
(ii) When n = 3k + 1:
n is not divisible by 3.
n + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1)  n + 2 is divisible by 3.
n + 4 = (3k + 1) + 4 = 3k + 5 = 3(k + 1) + 2  n + 4 is not divisible by 3.
(iii) When n = 3k + 2:
n is not divisible by 3.
n + 2 = (3k + 2) + 2 = 3k + 4 = 3(k + 1) + 1  n + 2 is not divisible by 3.
n + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2)  n + 4 is divisible by 3.
Hence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.
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In this case one out of n,n+2or n+4 is divisible by 3
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