View source for 2019 AMC 10C Problems/Problem 23

===Problem===
Bernado has an infinite amount of red, blue, orange, pink, yellow, purple, and black blocks. He puts them in the 2 by 2019 grid such that adjacent blocks are of different colors. What is the hundreds digit of the number of ways he can put the blocks in?


<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math>

===Solution===

There are <math>7 \times 6 = 42</math> choices to choose the colors for first <math>1 \times 2</math> section. Without loss of generality, assume that the first group of two is Red-Blue. There are <math>6^2 = 36</math> ways to choose the colors of the next <math>1 \times 2</math> section, but we would over-count by 5 if the new top and bottom are the same.Thus, there are only <math>36-5=31</math> distinct choices for the second group of vertical blocks. Now we just have to find the hundreds digit of <math>42 \times 31^{2018}</math>.
Now we can do some binomial expansions to find the hundreds digit of that number:
<math>42\cdot 31^{2018} \pmod {1000} </math>
<math>42 \cdot (3 \cdot 10 +1 )^{2018} \pmod {1000}</math>
<math>42 \cdot (1^{2018}+ 30 \cdot1^{2017} + 30^2 \cdot 1^{2016}...) \pmod {1000}</math>
<math>42 \cdot (1 + 30 + 900) \pmod {1000}</math>
<math>42 \cdot (931) \pmod {1000}</math>
<math> 102 \pmod {1000}</math>
Thus, the answer is <math>1</math>.

==Video Solution==
https://youtu.be/U3v0LeuA9SA

~IceMatrix