On a board measuring 1×100, each square is numbered from 1 to 100. Three colors are used to paint the squares from left to right. The following pattern is repeated: 1 blue square, 2 red squares, 3 green squares, 1 blue, 2 red, 3 green, and so on. What is the highest numbered square that is painted blue?
The correct answer and explanation is:
We are given a 1×100 board with squares numbered from 1 to 100, painted in a repeating pattern:
- 1 blue
- 2 red
- 3 green
This pattern repeats indefinitely: 1 blue, 2 red, 3 green, 1 blue, 2 red, 3 green, etc.
Step 1: Understand the pattern
The pattern covers 1 + 2 + 3 = 6 squares in each full cycle:
- Blue at position 1
- Red at positions 2 and 3
- Green at positions 4, 5, and 6
This 6-square cycle repeats across the board.
Step 2: Determine how many full cycles fit into 100 squares
We divide 100 by 6: 100÷6=16 full cycles with a remainder of 4100 \div 6 = 16 \text{ full cycles with a remainder of } 4
So:
- 16 full 6-square patterns = 16 × 6 = 96 squares
- Remaining: 100 − 96 = 4 squares
Step 3: Count how many blue squares appear
Each full cycle has exactly 1 blue square, and the blue square is always the first square in the cycle.
So:
- From 16 full cycles → 16 blue squares
- The 17th cycle begins at square 97
Now check the remainder:
- Square 97: blue
- Square 98: red
- Square 99: red
- Square 100: green (if needed, though only 4 squares remain in this partial cycle)
Therefore, square 97 is also blue.
Final count: 16 (from full cycles) + 1 (from the start of the partial cycle) = 17 blue squares
The positions of blue squares occur every 6 squares, starting at 1: 1,7,13,19,…,971, 7, 13, 19, …, 97
This forms an arithmetic sequence:
- First term: 1
- Common difference: 6
- General term: an=1+(n−1)⋅6=6n−5a_n = 1 + (n – 1) \cdot 6 = 6n – 5
Set 6n−5≤1006n – 5 \leq 100: 6n≤105⇒n≤17.56n \leq 105 \Rightarrow n \leq 17.5
So the maximum integer n=17n = 17 a17=6(17)−5=102−5=97a_{17} = 6(17) – 5 = 102 – 5 = \boxed{97}