How to go about this problem?
$\displaystyle \displaystyle 7^{2010}\equiv x \ \mbox{(mod 100)}$
$\displaystyle \displaystyle 7^5\equiv 07 \ \mbox{(mod 100)}$
$\displaystyle \displaystyle (7^{5})^{402}\equiv 7^{402}\equiv 7^{400+2}\equiv 7^{400}*7^2\equiv (7^5)^{80}*49\equi 7^{80}*49 \ \mbox{(mod 100)}$
Now as Soroban said $\displaystyle 7^4\equiv 1 \ \mbox{mod 100}$
$\displaystyle \displaystyle 7^{80}*49\equiv (7^4)^{20}*49\equiv 1^{20}*49\equiv 49 \ \mbox{(mod 100)}$
Sorry about the mistake
I'm not sure I get the above: the powers of 7 have a periodicy length of 4, and all the powers of 7 which are 2 mod 4 end with 49, so
the difference must be 9-4 = 5 ...Is this what you meant?
Also, why to write $\displaystyle 7^5 = 07\!\!\pmod {100}$ and not simply $\displaystyle 7^5=7\!\!\pmod {100}$ ?
Tonio
Hello, pranay!
$\displaystyle \text{Find the difference between the last two digits of }7^{2010}$
We have: .$\displaystyle 7^4 \;=\; 2401 \quad\Rightarrow\quad 7^4 \;\equiv\; 1\text{ (mod 100)}$
Then: .$\displaystyle 7^{2010} \;=\; 7^{4(502) + 2} \;=\; 7^{4(502)}\cdot7^2 \;=\;(7^4)^{502}\cdot49 $
Hence: .$\displaystyle (7^4)^{502}\cdot49 \;\equiv\; 1^{502}\cdot49 \text{ (mod 100)} \;\equiv\; 49 \text{ (mod 100)}$
The difference is: .$\displaystyle |4-9| \:=\:5$