1. ## 9^103

Find the sum of the last 2 digits of 9^103

What is the product of the last 2 digits of 2004^2004

What other ways to do these besides Euler's theorem and finding the pattern?

if you use factorials for (a-b)^103 you can write the equation back to front and add it together to get twice the value but cancelling out the intermediate factorials then divde the equation by half to get the number itself then just add the last two digits of it.

Someone explain it?

2. Use the binomial theorem.

$\displaystyle 9^{103}= (10- 1)^{103}$. By the binomial theorem, the last two terms of $\displaystyle (x- y)^{103}$ are $\displaystyle \left(\begin{array}{c}103 \\ 102\end{array}\right)xy^{102}+ \left(\begin{array}{c}103 \\ 103\end{array}\right)y^{103}= 103xy^{102}+ y^{103}$. The last two terms of this particular binomial expansion are 103(10)(1)+ (-1)= 102. All higher terms involve integers times 100 or higher so the last two digits are "02".

Similarly $\displaystyle 2004^{2004}= (2000+ 4)^{2004}$. Now the last two terms are $\displaystyle 2004(2000)4^{2003}+ 4^{2004}$ so the problem reduces to finding the last two digits of $\displaystyle 4^{2004}$.

3. How do I calculate 4^2004 using the same method? How do I break the 4 down? If I break it down to 3+1 or 5-1, there's a 1 there. And multiples of 4 don't have 1 as their last digit