SOLUTION: A bag contains red and blue tiles. Each tile has a number from the set \{-1, 0, 1\} written on it. I want to arrange 7 of these tiles in a row, so that the numbers on any three con

Question 1210216: A bag contains red and blue tiles. Each tile has a number from the set \{-1, 0, 1\} written on it. I want to arrange 7 of these tiles in a row, so that the numbers on any three consecutive tiles sum to 3. In how many ways can this be done, assuming that there are an unlimited number of tiles for any color and number combination?

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Solution:
Let the sequence of 7 tiles be $t_1, t_2, t_3, t_4, t_5, t_6, t_7$.
The number on each tile is from the set $\{-1, 0, 1\}$.
The condition is that the sum of the numbers on any three consecutive tiles is 3.
Consider the first three consecutive tiles:
$t_1 + t_2 + t_3 = 3$
Since each tile has a number from $\{-1, 0, 1\}$, the maximum sum of three consecutive tiles is $1 + 1 + 1 = 3$.
This implies that for the sum to be 3, each of the three consecutive tiles must have the number 1.
So, $t_1 = 1, t_2 = 1, t_3 = 1$.
Now consider the next three consecutive tiles:
$t_2 + t_3 + t_4 = 3$
Substituting the values of $t_2$ and $t_3$:
$1 + 1 + t_4 = 3$
$2 + t_4 = 3$
$t_4 = 1$
Continuing this pattern for all consecutive triplets:
$t_3 + t_4 + t_5 = 3 \implies 1 + 1 + t_5 = 3 \implies t_5 = 1$
$t_4 + t_5 + t_6 = 3 \implies 1 + 1 + t_6 = 3 \implies t_6 = 1$
$t_5 + t_6 + t_7 = 3 \implies 1 + 1 + t_7 = 3 \implies t_7 = 1$
The only possible sequence of numbers on the 7 tiles that satisfies the given condition is $(1, 1, 1, 1, 1, 1, 1)$.
Now, we need to consider the colors of the tiles. Each of the 7 tiles can be either red or blue. Since there are an unlimited number of tiles for any color and number combination, the color of each tile is independent of the number and the colors of other tiles.
For each of the 7 positions in the row, there are 2 choices for the color (red or blue).
The total number of ways to assign colors to the 7 tiles is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7 = 128$.
Since the only possible sequence of numbers is $(1, 1, 1, 1, 1, 1)$, we just need to find the number of ways to color these 7 tiles.
The number of ways to arrange 7 tiles with the number 1 such that the sum of any three consecutive tiles is 3 is 1 (the sequence must be all 1s).
For this unique numerical sequence, there are $2^7 = 128$ possible color combinations.
Final Answer: The final answer is $\boxed{128}$
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