Question

\[\left( {{n^3} - n} \right)\] is divisible by 3 .Explain the reason.

Answer 1

Hint: In this question, we will use the theorem of Euclid’s division lemma .By applying division lemma we will get to know that for any positive integer in the form of 3q$n$, 3q+1 or 3q+2 is divisible by 3.

Complete step-by-step solution -
We are given that $\left( {{n^3} - n} \right)$ is divisible by 3, it can also be written as:
$ \Rightarrow $$n\left( {{n^2} - 1} \right)$= $n\left( {n + 1} \right)\left( {n - 1} \right)$
Whenever a number is divided by 3 ,the remainder obtained is either 0,1 or 2.
 $\therefore $ $n$= $3p$ or $3p + 1$ or $3p + 2$,where p is some integer.
Ca$n$se1:
 If =, then is divisible by 3.
Case2:
 If $n = 3p + 1$, then $n - 1 = 3p + 1 - 1 = 3p$ is divisible by 3.
Cae3:
 if $n = 3p + 2$, then $n + 1 = 3p + 2 + 1 = 3p + 3 = 3\left( {p + 1} \right)$ is divisible by 3.
Let us take an example , put $n = 5$,
$ \Rightarrow \left( {{n^3} - n} \right) = \left( {{5^3} - 5} \right) = \left( {125 - 5} \right) = 120$.
Here ,if we do factors of 120, then we get
 $
   \Rightarrow 120 = 5 \times 6 \times 4 = 20 \times 6 \\
   \Rightarrow 120 = 3\left( {40} \right) \\
$
So here we can clearly see that 120 is a multiple of 3,
And 120 is also divisible by 3.
Hence, from the above three cases, we have seen that one of the numbers among $n,\left( {n + 1} \right),\left( {n - 1} \right)$ is always divisible by 3.
$ \Rightarrow n\left( {n + 1} \right)\left( {n - 1} \right)$=$\left( {{n^3} - n} \right)$ is divisible by 3.

Note :- In such types of problems we have to make cases so that we can easily identify which number is divisible or which is not divisible. Then by putting some random integers values we will find that either the number is divisible or not.