Question

The last digit of the number ${{3}^{400}}$.
\[\begin{align}
  & A.3 \\
 & B.9 \\
 & C.7 \\
 & D.1 \\
\end{align}\]

Answer 1

Hint: In this question, we need to find the last digit of the number ${{3}^{400}}$. For this we will use the concept of congruence modulo. If a is divided by n and b is the remainder we can write this form as $a\equiv b\bmod n$ (a is congruent to b mod n). We will divide ${{3}^{400}}$ by n to get the remainder which will be our last digit. We will use following properties of congruent modulo:
If $a\equiv b\bmod n$ then ${{a}^{m}}\equiv {{b}^{m}}\bmod n$ where m is any integer.

Complete step by step answer:
Here we are given the number as ${{3}^{400}}$. We need to find its last digit. As we know for a number, if we divide it by 10, the remainder will give us its last digits. For example, divide 124 by 10,
\[10\overset{12}{\overline{\left){\begin{align}
  & 124 \\
 & 10\downarrow \\
 & \overline{\begin{align}
  & 024 \\
 & 020 \\
 & \overline{004} \\
\end{align}} \\
\end{align}}\right.}}\]
Here the remainder is 4 which is the last digit so we will use this.
Let us use congruence modulo to find remainder when ${{3}^{400}}$ is divided by 10. We know $a\equiv b\bmod n$ represents that b is the remainder when a is divided by n.
So we need to find x for which ${{3}^{400}}\equiv x\bmod 10\cdots \cdots \cdots \left( 1 \right)$.
We know that 81 when divided by 10 gives us a remainder as 1. Therefore, we can write $81\equiv 1\bmod 10$.
Now $81=3\times 3\times 3\times 3={{3}^{4}}$ so we get ${{3}^{4}}\equiv 1\bmod 10$.
By property of congruence modulo, if $a\equiv b\bmod n$ then ${{a}^{m}}\equiv {{b}^{m}}\bmod n$ where m is an integer. So taking m as 100 here ${{\left( {{3}^{4}} \right)}^{100}}\equiv {{\left( 1 \right)}^{100}}\bmod 10$.
Using ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\text{ and }{{\left( 1 \right)}^{n}}=1$ we get ${{3}^{400}}\equiv 1\bmod 10\cdots \cdots \cdots \left( 2 \right)$.
Comparing (1) with (2) we get x = 1.
When ${{3}^{400}}$ is divided by 10, we get remainder as 1. Hence 1 is the last digit of ${{3}^{400}}$.

So, the correct answer is “Option D”.

Note: Students should know the properties of congruence modulo to solve this sum. Students can also find the solution of this sum using binomial expansion. They could use \[{{3}^{400}}={{\left( {{3}^{4}} \right)}^{100}}={{\left( 81 \right)}^{100}}={{\left( 1+80 \right)}^{100}}\] and then apply binomial expansion. From the first two terms they can say that the last digit is 1.