
Find the last digit of ${{3}^{57}}$.
Answer
605.1k+ views
Hint: For solving the problem related to finding the last digit of a large number, we start by taking small cases. Then we develop a familiar pattern which helps us to find the answer to the larger problem. In our case, we start by finding the last digit of 3, ${{3}^{2}}$ , ${{3}^{3}}$ and so on. Then by developing a familiar pattern we can find the last digit of ${{3}^{57}}$.
Complete step-by-step answer:
To solve this question, first start by analysing few of the powers of 3 and their last digits.
$\begin{align}
& {{3}^{1}}=3 \\
& {{3}^{2}}=9 \\
& {{3}^{3}}=27 \\
& {{3}^{4}}=81 \\
& {{3}^{5}}=243 \\
& {{3}^{6}}=729 \\
& {{3}^{7}}=2187 \\
& {{3}^{8}}=6561 \\
& {{3}^{9}}=19683 \\
\end{align}$
We can observe a familiar pattern from these powers of 3. We see that the last digits of these powers get repeated after ${{3}^{4}}$. Thus, the patterns of these last digits are 3, 9, 7, 1, 3, 9, 7, 1 and so on. Thus, we can confirm that only these four numbers (3, 9, 7, 1) will appear as the last digit in any power of 3. Further, we see that pattern repeats after four numbers. Now, having this information, let’s try to utilize this information to find a connection to ${{3}^{57}}$. We will try to find the remainder of the number of the power when divided by 4 (since, the pattern repeats after 4 numbers). Thus, say, we want to find the last digit of ${{3}^{6}}$ , we will find the remainder when 6 is divided by 4. We get the remainder as 2. Now, we can see that ${{3}^{6}}$ has the same last digit as three raised to the power of the remainder when 6 is divided by 4 (that is 2). Thus, we can use this result in the case of ${{3}^{57}}$ , we have, when 57 is divided by 4, the remainder is 1. Thus, the last digit will be the same as that of ${{3}^{1}}$.
Hence, the last digit of ${{3}^{57}}$ is 3.
Note: In the above problem, we used the technique of finding the last digit of the number with the help of repeated patterns. There are also other techniques like applying modular arithmetic, Chinese remainder theorem, etc. which can be used to find the last digit of a number (although these techniques require knowledge of higher level mathematics).
Complete step-by-step answer:
To solve this question, first start by analysing few of the powers of 3 and their last digits.
$\begin{align}
& {{3}^{1}}=3 \\
& {{3}^{2}}=9 \\
& {{3}^{3}}=27 \\
& {{3}^{4}}=81 \\
& {{3}^{5}}=243 \\
& {{3}^{6}}=729 \\
& {{3}^{7}}=2187 \\
& {{3}^{8}}=6561 \\
& {{3}^{9}}=19683 \\
\end{align}$
We can observe a familiar pattern from these powers of 3. We see that the last digits of these powers get repeated after ${{3}^{4}}$. Thus, the patterns of these last digits are 3, 9, 7, 1, 3, 9, 7, 1 and so on. Thus, we can confirm that only these four numbers (3, 9, 7, 1) will appear as the last digit in any power of 3. Further, we see that pattern repeats after four numbers. Now, having this information, let’s try to utilize this information to find a connection to ${{3}^{57}}$. We will try to find the remainder of the number of the power when divided by 4 (since, the pattern repeats after 4 numbers). Thus, say, we want to find the last digit of ${{3}^{6}}$ , we will find the remainder when 6 is divided by 4. We get the remainder as 2. Now, we can see that ${{3}^{6}}$ has the same last digit as three raised to the power of the remainder when 6 is divided by 4 (that is 2). Thus, we can use this result in the case of ${{3}^{57}}$ , we have, when 57 is divided by 4, the remainder is 1. Thus, the last digit will be the same as that of ${{3}^{1}}$.
Hence, the last digit of ${{3}^{57}}$ is 3.
Note: In the above problem, we used the technique of finding the last digit of the number with the help of repeated patterns. There are also other techniques like applying modular arithmetic, Chinese remainder theorem, etc. which can be used to find the last digit of a number (although these techniques require knowledge of higher level mathematics).
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