What is the remainder when $\left((7)^{77}\right)^{77}$ is divided by 10 ?

Question

Professor PC Thomas Classes · Answer

Check video solution for simple method to solve this question
Let $\boldsymbol{N}=\left((7)^{77}ight)^{77}$. We need to find the remainder when $\boldsymbol{N}$ is divided by 10 , which is equivalent to finding the last digit of $\boldsymbol{N}$. We will use the concept of cyclicity of the last digits of powers.
First, let's observe the pattern of the last digits of powers of 7:
$7^1=7 \Longrightarrow$ last digit is 7
$7^2=49 \Longrightarrow$ last digit is 9
$7^3=343 \Longrightarrow$ last digit is 3
$7^4=2401 \Longrightarrow$ last digit is 1
$7^5=16807 \Longrightarrow$ last digit is 7
The pattern of the last digits of powers of 7 is $7,9,3,1$ which repeats every 4 powers. The cycle length is 4 .
To find the last digit of $7^E$, we need to determine the remainder of the exponent $E$ when divided by the cycle length (which is 4). In this problem, the exponent is $E=77^{77}$.
We need to calculate $77^{77}(\bmod 4)$.
First, let's find the remainder of the base of the exponent, 77 , when divided by 4 :
$$
77=4 	imes 19+1
$$
So, $77=1(\bmod 4)$.
Now, we can substitute this into the exponent calculation:
$77^{77}=1^{77}(\bmod 4)$
$77^{77}=1(\bmod 4)$.
This result tells us that the exponent $77^{77}$ can be written in the form $\boldsymbol{4} \boldsymbol{k}+\mathbf{1}$ for some integer $\boldsymbol{k}$.
Therefore, the last digit of $\left((7)^{7 \pi}ight)^{7 \pi}$ will be the same as the last digit of $7^1$, which is 7 .
Thus, when $\left((7)^{T 7}ight)^{T 7}$ is divided by 10 , the remainder is 7 .