Question

The remainder when ${{\left( 2017 \right)}^{2018}}+{{\left( 2018 \right)}^{2019}}+{{\left( 2019 \right)}^{2020}}$ is divided by $5$ is
$\begin{align}
  & \left( A \right)2 \\
 & \left( B \right)4 \\
 & \left( C \right)3 \\
 & \left( D \right)0 \\
\end{align}$

Answer 1

Hint: We work on the unit’s place digit for these problems. At first, we find the cycle for the powers of the numbers, that is, the power after which the unit’s place digit repeats itself. We then express the power as an integer multiple of that cycle and then find the unit’s place digit for that term. We then find all the unit’s place digit for all the terms and dividing the sum by $5$ gives us the remainder.

Complete step by step answer:
Since $5$ is a single digit number, therefore only the unit’s place digit of the entire expression will matter in finding out the remainder . the given expression is
${{\left( 2017 \right)}^{2018}}+{{\left( 2018 \right)}^{2019}}+{{\left( 2019 \right)}^{2020}}$
Let us first start with ${{\left( 2017 \right)}^{2018}}$ . The unit’s place digit of $2017$ is $7$ . Every digit has a cyclic behaviour of powers, that is, the unit’s place digit repeats itself after each definite interval. For example,
$\begin{align}
  & {{2}^{1}}=2 \\
 & {{2}^{2}}=4 \\
 & {{2}^{3}}=8 \\
 & {{2}^{4}}=16 \\
 & {{2}^{5}}=32 \\
\end{align}$
We can see that after the fourth power, the unit’s place digit becomes $2$ again. So, the cycle of $2$ is $4$ . If we express the power in the form $4n+a$ , then $a$ determines the unit’s place digit.
Similarly,
$\begin{align}
  & {{7}^{1}}=2 \\
 & {{7}^{2}}=49 \\
 & {{7}^{3}}=343 \\
 & {{7}^{4}}=2401 \\
 & {{7}^{5}}=16807 \\
\end{align}$
Thus, the cycle of $7$ is $4$ . We express the power of $2017$ , which is $2018$ as $4\left( 504 \right)+2$ . We see that the remainder being $2$ , all that matters for the unit’s place digit is this power $2$ . ${{7}^{2}}=49$ , which has the unit’s place digit $9$ . Thus, the unit’s place digit of ${{\left( 2017 \right)}^{2018}}$ is $9$ .
Proceeding as above for ${{\left( 2018 \right)}^{2019}}$ , the cycle for $8$ is also $4$ . We express the power of $2018$ , which is $2019$ as $4\left( 504 \right)+3$ . We see that the remainder being $3$ , all that matters for the unit’s place digit is this power $3$ . ${{8}^{3}}=512$ , which has the unit's place digit $2$ . Thus, the unit’s place digit of ${{\left( 2018 \right)}^{2019}}$ is $2$ .
Proceeding as above for ${{\left( 2019 \right)}^{2020}}$ , the cycle for $9$ is $2$ . We express the power of $2019$ , which is $2020$ as $2\left( 1010 \right)+0$ . We see that the remainder being $0$ , all that matters for the unit’s place digit is this power $0$ . ${{9}^{0}}=1$ , which has the unit's place digit $1$ . Thus, the unit’s place digit of ${{\left( 2019 \right)}^{2020}}$ is $1$ .
The sum of the unit’s place digit for the entire expression is thus,
$9+2+1=12$
Since, $12$ is greater than $5$ , we cannot say that the remainder is $12$ . So, we need to again divide this $12$ by $5$ which gives $2$ as remainder.
Therefore, we can conclude that the remainder when ${{\left( 2017 \right)}^{2018}}+{{\left( 2018 \right)}^{2019}}+{{\left( 2019 \right)}^{2020}}$ is divided by $5$ is $2$ .

Note: For these types of problems, we should always focus on the unit’s place digit and nothing else, as they don’t matter. The cycle of the powers of the numbers must be calculated carefully, as students make mistakes over here. At the end, we must be careful that the remainder that we are claiming is less than the divisor; if it is not, we should again divide it by the divisor and then find the remainder.