Question

Show that only one of the numbers \[n\], \[n + 2\] and \[n + 4\] is divisible by 3.

Answer 1

Hint:
We know that any positive integer of the form \[3q\] or, \[3q + 1\] or \[3q + 2\] for some integer q and one and only one of these possibilities can occur. Here we need to prove for \[n\], \[n + 2\] and \[n + 4\] in which by applying \[n = 3q\], \[n = 3q + 1\] and \[n = 3q + 2\] then simplify the terms to find out all the possibilities when it is divisible by 3.

Complete step by step solution:
Any number of the form of \[3q\], \[3q + 1\]or \[3q + 2\], we have following cases:
Case I: When \[n = 3q\]
In this case we have,
\[n = 3q\], which is divisible by 3,
Now again, \[n = 3q\]
\[n + 2 = 3q + 2\]
Hence, \[n + 2\] leaves remainder 2 when divided by 3 hence, \[n + 2\] is not divisible by 3.
Now again, \[n = 3q\]
\[n + 4 = 3q + 4 = 3\left( {q + 1} \right) + 1\]
Hence, \[n + 4\] leaves remainder 1 when divided by 3 hence, \[n + 4\] is not divisible by 3.
Thus, we can say that \[n\] is divisible by 3 but \[n + 2\] and n+4 is not divisible by 3.
Case II: When \[n = 3q + 1\]
In this case we have,
\[n = 3q + 1\]Here, \[n\] is not divisible by 3 as n leaves remainder 1.
Now, \[n = 3q + 1\]
\[n + 2 = \left( {3q + 1} \right) + 2 = 3q + 3 = 3\left( {q + 1} \right)\]
Hence, \[n + 2\]is divisible by 3.
Again \[n = 3q + 1\]

\[n + 4 = 3q + 1 + 4 = 3q + 5 = 3\left( {q + 1} \right) + 2\]
Hence, \[n + 4\] leaves remainder 2 when divided by 3, hence it is not divisible by 3.
Case III: When \[n = 3q + 2\]
In this case we have,
 \[n = 3q + 2\]
Here, \[n\] is not divisible by 3 as it leaves remainder 2.
Now, \[n = 3q + 2\]
\[n + 2 = 3q + 2 + 2 = 3q + 4 = 3\left( {q + 1} \right) + 1\]
Here, \[n + 2\] leaves remainder 1 when divided by 3 hence, it is not divisible by 3.
Again, \[n = 3q + 2\]
\[n + 4 = 3q + 2 + 4 = 3q + 6 = 3\left( {q + 2} \right)\]
Here, \[n + 4\] is divisible by 3.
Therefore, \[n + 4\] is divisible by 3 but \[n\] and \[n + 2\] is not divisible by 3.
Thus, one and only one out of n, \[n + 2\], \[n + 4\] is divisible by 3.

Note:
The key point to prove the number \[n\], \[n + 2\] and \[n + 4\] divisible by 3 is that we need to consider the \[n = 3q\], \[n = 3q + 1\] and \[n = 3q + 2\] for the given values with respect to \[n\] and a number is said to be divisible if does not contain remainder, if remainder exists then it is not divisible by the given number.