Question

Prove that one of every three consecutive positive integers is divisible by 3.

Answer 1

Hint: For solving this question we will see the mathematical result of the Euclid Division Lemma and then we will consider $\left( n-1 \right),n,\left( n+1 \right)$ as three consecutive positive integers. After that, we will consider three cases for $n$ like if $n$ is divisible by 3 or it leaves remainder 1 or 2 when it is divided by 3. Moreover, we will consider numerical examples and ultimately prove the desired result.

Complete step by step answer:
We have to prove that one of every three consecutive positive integers is divisible by 3.
Now, before we proceed we should know that if $a$ is a positive integer and when we divide another positive integer $b$ by $a$ and it leaves remainder $r$ where, $rNow, let $\left( n-1 \right),n,\left( n+1 \right)$ be three consecutive positive integers.
Now, there will be the following three cases:
First case:
Now, if $n$ is divisible by 3 so, we can write $n=3k$ where $k$ is any integer. Then,
$\begin{align}
  & n=3k \\
 & \Rightarrow n+1=3k+1 \\
 & \Rightarrow n-1=3k-1 \\
\end{align}$
Now, from the above result, we conclude that $n+1=3k+1$ which means $n+1$ is not divisible by 3 and as $n-1=3k-1$ which means $n-1$ is not divisible by 3. For example: if $n=3$ then, $n+1=4$ is not divisible by 3 and $n-1=2$ is not divisible by 3.
Second case:
Now, if $n$ is divided by 3 and it leaves remainder 1 so, we can write $n=3k+1$ where $k$ is any integer. Then,
$\begin{align}
  & n=3k+1 \\
 & \Rightarrow n+1=3k+2 \\
 & \Rightarrow n-1=3k \\
\end{align}$
Now, from the above result, we conclude that $n+1=3k+2$ which means $n+1$ is not divisible by 3 and as $n-1=3k$ which means $n-1$ is divisible by 3. For example: if $n=4$ , $n+1=5$ and $n-1=3$ . Then,

$3\overset{1}{\overline{\left){\begin{align}
  & 4 \\
 & \underline{3} \\
 & \underline{1} \\
\end{align}}\right.}}\text{ }3\overset{1}{\overline{\left){\begin{align}
  & 5 \\
 & \underline{3} \\
 & \underline{2} \\
\end{align}}\right.}}\text{ }3\overset{1}{\overline{\left){\begin{align}
  & 3 \\
 & \underline{3} \\
 & \underline{0} \\
\end{align}}\right.}}$
Now, it is evident that $n-1$ is divisible by 3 if $n$ leaves remainder 1 when divided by 3.
Third case:
Now, if $n$ is divided by 3 and it leaves remainder 2 so, we can write $n=3k+2$ where $k$ is any integer. Then,
$\begin{align}
  & n=3k+2 \\
 & \Rightarrow n+1=3k+3 \\
 & \Rightarrow n+1=3\left( k+1 \right) \\
 & \Rightarrow n-1=3k+1 \\
\end{align}$
Now, from the above result, we conclude that $n+1=3\left( k+1 \right)$ which means $n+1$ is divisible by 3 and as $n-1=3k+1$ which means $n-1$ is not divisible by 3. For example: if $n=8$ , $n+1=9$ and $n-1=7$ . Then,
$3\overset{2}{\overline{\left){\begin{align}
  & 8 \\
 & \underline{6} \\
 & \underline{2} \\
\end{align}}\right.}}\text{ }3\overset{3}{\overline{\left){\begin{align}
  & 9 \\
 & \underline{9} \\
 & \underline{0} \\
\end{align}}\right.}}\text{ }3\overset{2}{\overline{\left){\begin{align}
  & 7 \\
 & \underline{6} \\
 & \underline{1} \\
\end{align}}\right.}}$
Now, it is evident that $n+1$ is divisible by 3 if $n$ leaves remainder 2 when divided by 3.
Now, from the above result, we conclude that one of every three consecutive positive integers will be divisible by 3.
Hence, proved.

Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to prove the desired result. After that, we should apply the result of the Euclid Division Lemma correctly and use proper numerical examples for better understanding, then proceed stepwise without any confusion. Moreover, we should apply our intelligence while solving such problems and for better clarity of the concept we should take examples to strengthen our concepts of numerical systems.