Problem:
Find the last $2$ digits of $7^{2008}$.
Unfortunately I have no idea how to solve this problem. I know that for the last digit, we have to apply the concept of cyclicity, but I'm not aware of how to extend this to the last $2$ digits. I would be truly grateful for any help. Many thanks in advance!
On
$$7^4\equiv 01\pmod {100}\\\implies (7^4)^{502}\equiv (01)^{502}\pmod {100}\\\implies 7^{2008}\equiv 01\pmod {100} $$.
Hence the last two digits are $01$.
On
Hint: $~7^{2008}=49^{1004}=(50-1)^{1004}.~$ Now expand using the binomial theorem, and notice that all terms except for the first two are multiples of $50^2$, and therefore of $100$.
On
Since $7$ and $100$ are coprime with respect to each other, you can use Euler's theorem:
On
Start by listing some cases of $7^n \mod 100:$ $$7^0=\color{red}{01}$$ $$7^1=\color{red}{07}$$ $$7^2=\color{red}{49}$$ $$7^3=3\color{red}{43}$$ $$7^4=24\color{red}{01}$$ $$7^5=168\color{red}{07}$$ $$7^6=1176\color{red}{49}$$ $$7^7=8235\color{red}{43}$$ $$7^8=57648\color{red}{01}$$ $$7^9=403536\color{red}{07}$$ $$7^{10}=2824752\color{red}{49}$$ $$7^{11}=19773267\color{red}{43}$$ $$7^{12}=138412872\color{red}{01}$$ $$\dots$$
Do you see the pattern?
Just a clarification to Subhadeep's answer in case it is confusing.
Let us first build a small table with everything larger than 100 in parentheses: $$\begin{array}{lr}7^0 = &1\\ 7^1 =& 7\\ 7^2 =& 49\\ 7^3 =& (3)43\\ 7^4 =& (24)01\\ \end{array}$$
So we see that after 4 we get "back" to 1. This means $7^{4k}$ will end in 01 for all $k$. So if we can calculate the exponent modulo 4 we can see which of 01,07,49,43 it will be.
As the exponent 2008 is divisible by 4 it must be "01".