To find the last digit of an expression which is in the format of ab we have to learn the concept of cyclicity.
Digits 0, 1, 5, 6:
These digits always give the same units digit.
${0^n} = 0; \, {1^n} = 1; \, {5^n} = 5; \, {6^n} = 6$
Digits 4, 9:
For 4, If the power is odd you will get $4$ as units digit. If the power is even, you will get $6$ as units digit.
$\begin{array}{*{20}{c}}
{{4^{odd}} = 4}\\
{{4^{even}} = 6}
\end{array}$
For 9, If the power is odd you will get $9$ as units digit. If the power is even, you will get $1$ as units digit.
$\begin{array}{*{20}{c}}
{{9^{odd}} = 9}\\
{{9^{even}} = 1}
\end{array}$
Digits 2, 3, 7, 8:
Digits 2, 3, 7, 8 follows cyclicity of $4$.
For example, 21 = 2, 22 = 4, 23 = 8, 24 = 6, 25 = 2, 26 = 4 and the cycle repeats.
Shortcut For numbers 2, 3, 7, 8, find the remainder when the power is divided by 4.
If the remainder is 0, then replace the original power with $4$.
For example, ${2^{100}}$. The power is a multiple of 4. So ${2^{100}} = {2^4} = 6$
If the remainder is $1,\, 2,\,3$ then take the remainder as power.
For example, ${7^{50}} = {7^2} = 9$
Detailed Concept:
Case1: If the power if multiple of 4, then take the power as 4.
${2^{4n}} = {\left( {{2^4}} \right)^n} = {6^n} = 6$
${3^{4n}} = {\left( {{3^4}} \right)^n} = {1^n} = 1$
${7^{4n}} = {\left( {{7^4}} \right)^n} = {1^n} = 1$
${8^{4n}} = {\left( {{8^4}} \right)^n} = {6^n} = 6$
Case2: If the power not a multiple of 4, then find the remainder and take this as the original power.
If the remainder is $1$,
${2^{4n + 1}} = {\left( {{2^4}} \right)^n} \times {2^1} = {6^n} \times {2^1} = 2$ \(\quad \left( \because {6 \times 2 = 2} \right)\)
${3^{4n + 1}} = {\left( {{3^4}} \right)^n} \times {3^1} = {1^n} \times {3^1} = 3$
${7^{4n + 1}} = {\left( {{7^4}} \right)^n} \times {7^1} = {1^n} \times {7^1} = 7$
${8^{4n + 1}} = {\left( {{8^4}} \right)^n} \times {8^1} = {6^n} \times {8^1} = 8$
If the remainder is $2$,
${2^{4n + 2}} = {\left( {{2^4}} \right)^n} \times {2^2} = {6^n} \times {2^2} = 4$
${3^{4n + 2}} = {\left( {{3^4}} \right)^n} \times {3^2} = {1^n} \times {3^2} = 9$
${7^{4n + 2}} = {\left( {{7^4}} \right)^n} \times {7^2} = {1^n} \times {7^2} = 9$
${8^{4n + 2}} = {\left( {{8^4}} \right)^n} \times {8^2} = {6^n} \times {8^2} = 4$
If the remainder is $3$,
${2^{4n + 3}} = {\left( {{2^4}} \right)^n} \times {2^3} = {6^n} \times {2^3} = 8$
${3^{4n + 3}} = {\left( {{3^4}} \right)^n} \times {3^3} = {1^n} \times {3^3} = 7$
${7^{4n + 3}} = {\left( {{7^4}} \right)^n} \times {7^3} = {1^n} \times {7^3} = 3$
${8^{4n + 3}} = {\left( {{8^4}} \right)^n} \times {8^3} = {6^n} \times {8^3} = 2$
Let us put all the results in a table for quick reference:
Digits 0, 1, 5, 6:
These digits always give the same units digit.
${0^n} = 0; \, {1^n} = 1; \, {5^n} = 5; \, {6^n} = 6$
Digits 4, 9:
For 4, If the power is odd you will get $4$ as units digit. If the power is even, you will get $6$ as units digit.
$\begin{array}{*{20}{c}}
{{4^{odd}} = 4}\\
{{4^{even}} = 6}
\end{array}$
For 9, If the power is odd you will get $9$ as units digit. If the power is even, you will get $1$ as units digit.
$\begin{array}{*{20}{c}}
{{9^{odd}} = 9}\\
{{9^{even}} = 1}
\end{array}$
Digits 2, 3, 7, 8:
Digits 2, 3, 7, 8 follows cyclicity of $4$.
For example, 21 = 2, 22 = 4, 23 = 8, 24 = 6, 25 = 2, 26 = 4 and the cycle repeats.
Shortcut For numbers 2, 3, 7, 8, find the remainder when the power is divided by 4.
If the remainder is 0, then replace the original power with $4$.
For example, ${2^{100}}$. The power is a multiple of 4. So ${2^{100}} = {2^4} = 6$
If the remainder is $1,\, 2,\,3$ then take the remainder as power.
For example, ${7^{50}} = {7^2} = 9$
Detailed Concept:
Case1: If the power if multiple of 4, then take the power as 4.
${2^{4n}} = {\left( {{2^4}} \right)^n} = {6^n} = 6$
${3^{4n}} = {\left( {{3^4}} \right)^n} = {1^n} = 1$
${7^{4n}} = {\left( {{7^4}} \right)^n} = {1^n} = 1$
${8^{4n}} = {\left( {{8^4}} \right)^n} = {6^n} = 6$
Case2: If the power not a multiple of 4, then find the remainder and take this as the original power.
If the remainder is $1$,
${2^{4n + 1}} = {\left( {{2^4}} \right)^n} \times {2^1} = {6^n} \times {2^1} = 2$ \(\quad \left( \because {6 \times 2 = 2} \right)\)
${3^{4n + 1}} = {\left( {{3^4}} \right)^n} \times {3^1} = {1^n} \times {3^1} = 3$
${7^{4n + 1}} = {\left( {{7^4}} \right)^n} \times {7^1} = {1^n} \times {7^1} = 7$
${8^{4n + 1}} = {\left( {{8^4}} \right)^n} \times {8^1} = {6^n} \times {8^1} = 8$
If the remainder is $2$,
${2^{4n + 2}} = {\left( {{2^4}} \right)^n} \times {2^2} = {6^n} \times {2^2} = 4$
${3^{4n + 2}} = {\left( {{3^4}} \right)^n} \times {3^2} = {1^n} \times {3^2} = 9$
${7^{4n + 2}} = {\left( {{7^4}} \right)^n} \times {7^2} = {1^n} \times {7^2} = 9$
${8^{4n + 2}} = {\left( {{8^4}} \right)^n} \times {8^2} = {6^n} \times {8^2} = 4$
If the remainder is $3$,
${2^{4n + 3}} = {\left( {{2^4}} \right)^n} \times {2^3} = {6^n} \times {2^3} = 8$
${3^{4n + 3}} = {\left( {{3^4}} \right)^n} \times {3^3} = {1^n} \times {3^3} = 7$
${7^{4n + 3}} = {\left( {{7^4}} \right)^n} \times {7^3} = {1^n} \times {7^3} = 3$
${8^{4n + 3}} = {\left( {{8^4}} \right)^n} \times {8^3} = {6^n} \times {8^3} = 2$
Let us put all the results in a table for quick reference:
$\begin{array}{|c|c|c|c|}
\hline
Remainder \to \\ Base \downarrow
& 0 & 1 & 2 & 3\\
& (\because Power\; is \\
& multiple\; of\;4) \\\hline
2 & {\left( {{2^4}} \right)^n} = 6 & {2^1} = 2 & {2^2} = 4 & {2^3} = 8\\ \hline
3 & {\left( {{3^4}} \right)^n} = 1 & {3^1} = 3& {3^2} = 9 & {3^3} = 7\\ \hline
7 & {\left( {{7^4}} \right)^n} = 1 & {7^1} = 7& {7^2} = 9 & {7^3} = 3\\ \hline
8 & {\left( {{8^4}} \right)^n} = 8 & {8^1} = 8& {8^2} = 4 & {8^3} = 2\\ \hline
\end{array}$
\hline
Remainder \to \\ Base \downarrow
& 0 & 1 & 2 & 3\\
& (\because Power\; is \\
& multiple\; of\;4) \\\hline
2 & {\left( {{2^4}} \right)^n} = 6 & {2^1} = 2 & {2^2} = 4 & {2^3} = 8\\ \hline
3 & {\left( {{3^4}} \right)^n} = 1 & {3^1} = 3& {3^2} = 9 & {3^3} = 7\\ \hline
7 & {\left( {{7^4}} \right)^n} = 1 & {7^1} = 7& {7^2} = 9 & {7^3} = 3\\ \hline
8 & {\left( {{8^4}} \right)^n} = 8 & {8^1} = 8& {8^2} = 4 & {8^3} = 2\\ \hline
\end{array}$
