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?


$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

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

Video Solution

https://youtu.be/U3v0LeuA9SA

~IceMatrix

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