
The unit digit in the decimal expansion of ${{7}^{25}}$ is
A.1
B.3
C.5
D.7
Answer
574.5k+ views
Hint: We will evaluate the first few powers of 7. We know that the unit digits in the increasing power of any positive number repeats after a certain period of time. Here, we will notice the same behavior with the unit digits in the power of 7. The pattern followed by the unit digits will be 7, 9, 3, 1, 7, 9, 3, 1……
Complete step by step solution:
Here, we will take the help of congruence modulo i.e.
$a\equiv b\left( \bmod c \right)$, which means that we will get the same remainder when we divide a or b by c.
We will start evaluating the powers of 7.
${{7}^{1}}=7$, here the unit digit is 7i.e. ${{7}^{1}}\equiv 7\left( \bmod 10 \right)$
${{7}^{2}}=49$, here the unit digit is 7i.e. ${{7}^{2}}\equiv 9\left( \bmod 10 \right)$
${{7}^{3}}=343$, here the unit digit is 7i.e. ${{7}^{3}}\equiv 3\left( \bmod 10 \right)$
${{7}^{4}}=2401$, here the unit digit is 7i.e. ${{7}^{4}}\equiv 1\left( \bmod 10 \right)$
Thus, we have found that 7 is in unit place when 4k+1; 9 is in unit place when 4k+2; 3 is in unit place when 4k+3 and 1 is in unit place when 4k.
$\begin{align}
& {{7}^{1}}\equiv 7\left( \bmod 10 \right) \\
& {{7}^{2}}\equiv 9\left( \bmod 10 \right) \\
& {{7}^{3}}\equiv 3\left( \bmod 10 \right) \\
& {{7}^{4}}\equiv 1\left( \bmod 10 \right) \\
& {{7}^{4k+1}}\equiv 7\left( \bmod 10 \right) \\
& {{7}^{4k+2}}\equiv 9\left( \bmod 10 \right) \\
& {{7}^{4k+3}}\equiv 3\left( \bmod 10 \right) \\
& {{7}^{4k}}\equiv 1\left( \bmod 10 \right) \\
\end{align}$
We have $n=25$, therefore , the power is of type 4k+1. Thus, the unit digit will be 7.
Hence, the correct answer is option D.
Note: The digit in the unit place is found by taking mode of that number.
Therefore, if the digit in the unit place is m then it can be written as ${{7}^{n}}\equiv m\left( \bmod 10 \right)$
We have started evaluating the power of the first few powers of 7 and we have observed that the digit in the unit place is following a particular pattern i.e. the same digit is repeating after a certain period of time.
The pattern has a period of 4 as only four digits in the unit place are repeating again and again.
Complete step by step solution:
Here, we will take the help of congruence modulo i.e.
$a\equiv b\left( \bmod c \right)$, which means that we will get the same remainder when we divide a or b by c.
We will start evaluating the powers of 7.
${{7}^{1}}=7$, here the unit digit is 7i.e. ${{7}^{1}}\equiv 7\left( \bmod 10 \right)$
${{7}^{2}}=49$, here the unit digit is 7i.e. ${{7}^{2}}\equiv 9\left( \bmod 10 \right)$
${{7}^{3}}=343$, here the unit digit is 7i.e. ${{7}^{3}}\equiv 3\left( \bmod 10 \right)$
${{7}^{4}}=2401$, here the unit digit is 7i.e. ${{7}^{4}}\equiv 1\left( \bmod 10 \right)$
Thus, we have found that 7 is in unit place when 4k+1; 9 is in unit place when 4k+2; 3 is in unit place when 4k+3 and 1 is in unit place when 4k.
$\begin{align}
& {{7}^{1}}\equiv 7\left( \bmod 10 \right) \\
& {{7}^{2}}\equiv 9\left( \bmod 10 \right) \\
& {{7}^{3}}\equiv 3\left( \bmod 10 \right) \\
& {{7}^{4}}\equiv 1\left( \bmod 10 \right) \\
& {{7}^{4k+1}}\equiv 7\left( \bmod 10 \right) \\
& {{7}^{4k+2}}\equiv 9\left( \bmod 10 \right) \\
& {{7}^{4k+3}}\equiv 3\left( \bmod 10 \right) \\
& {{7}^{4k}}\equiv 1\left( \bmod 10 \right) \\
\end{align}$
We have $n=25$, therefore , the power is of type 4k+1. Thus, the unit digit will be 7.
Hence, the correct answer is option D.
Note: The digit in the unit place is found by taking mode of that number.
Therefore, if the digit in the unit place is m then it can be written as ${{7}^{n}}\equiv m\left( \bmod 10 \right)$
We have started evaluating the power of the first few powers of 7 and we have observed that the digit in the unit place is following a particular pattern i.e. the same digit is repeating after a certain period of time.
The pattern has a period of 4 as only four digits in the unit place are repeating again and again.
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