Question

If n leaves a remainder of 1 when divided by 3, then show that ${n^3}$ also leaves a remainder of 1 when divided by 3.
A. True
B. False

Answer 1

Hint: We will first assume n to be something as per the given data. Then, we will find its cube and term it as ${n^3}$. After that, we will club its terms such that, all the terms are multiple of 3 except 1 and thus it leaves a remainder of 1 as well.

Complete step-by-step solution:
Since, we are already given in the question that n leaves a remainder of 1 when divided by 3.
$\therefore n = 3m + 1$ for some m in Natural Numbers.
Now, taking the cube on both the sides, we will then get:-
$ \Rightarrow {n^3} = {\left( {3m + 1} \right)^3}$
Now, we will use the formula given by ${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$.
$ \Rightarrow {n^3} = {\left( {3m + 1} \right)^3} = {(3m)^3} + {1^3} + 3 \times 3m(3m + 1)$
Simplifying the RHS of the above expression, we will get:-
$ \Rightarrow {n^3} = 27{m^3} + 1 + 9m(3m + 1)$
Simplifying the RHS of the above expression further, we will get:-
$ \Rightarrow {n^3} = 27{m^3} + 1 + 27{m^2} + 9m$
We can write it as:-
$ \Rightarrow {n^3} = 3\left( {9{m^3} + 9{m^2} + 3m} \right) + 1$
Now, we can clearly see that except 1, all the terms in the cube of n are divisible by 3.
$\therefore {n^3}$ also leaves a remainder of 1 when divided by 3.

$\therefore $ The answer is (A) true.

Note: We can also use an alternate way to do the same. The alternate method is known as the method of contradiction. It will be very much similar but, we will just assume a contrary statement.
Let us assume that n leaves a remainder of 1 when divided by 3 but ${n^3}$ does not leave a remainder of 1 when divided by 3.
Since, we are already given that n leaves a remainder of 1 when divided by 3.
$\therefore n = 3m + 1$ for some m in Natural Numbers.
Now, go on exactly as we did in solution until we get:-
${n^3}$ also leaves a remainder of 1 when divided by 3.
Now, since we assumed that ${n^3}$ does not leave a remainder of 1 when divided by 3.
Hence, we have a contraction to the assumed statement.
$\therefore {n^3}$ also leaves a remainder of 1 when divided by 3.
$\therefore $ the answer is (A) true.