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Prove that one and only one out of $n, (n + 2)$ and $(n + 4)$ is divisible by 3, where $n$ is any positive integer.
To do:
We have to prove that one and only one out of $n, (n + 2)$ and $(n + 4)$ is divisible by 3, where $n$ is any positive integer.
Solution:
Let $a = n, b = n + 2$ and $c =n + 4$
Order triplet is $(a, b, c) = (n, n + 2, n + 4)$....…(i)
where, $n$ is any positive integer
At $n = 1$
$(a, b, c) = (1, 1 + 2, 1 + 4)$
$= (1, 3, 5)$
At $n = 2$
$(a, b, c) = (2, 2 + 2, 2 + 4)$
$= (2, 4, 6)$
At $n = 3$
$(a, b,c) = (3, 3 + 2, 3 + 4)$
$= (3,5,7)$
At $n =4$
$(a,b, c) =(4, 4 + 2, 4 +4)$
$= (4, 6, 8)$
At $n = 5$
$(a, b,c) = (5, 5 + 2, 5 +4)$
$= (5,7,9)$
At $n = 6$
$(a,b, c) = (6, 6 + 2, 6 + 4)$
$= (6,8,10)$
At $n = 7$
$(a, b,c) = (7, 7 + 2, 7 + 4)$
$= (7, 9,11)$
At $n = 8$
$(a, b,c) = (8,8+ 2, 8+ 4)$
$= (8,10,12)$
We observe that each triplet consists of one and only one number which is a multiple of 3.
Hence, one and only one out of $n, (n + 2)$ and $(n + 4)$ is divisible by 3, where, $n$ is any positive integer.