Question

a choir director is planning how she will arrange the choir members for an upcoming performance. she arranges them in rows that form a triangle and alternates the shirt color the members will wear between red and blue, to maximize sound and visual effect. a red shirt will always be in the front row. design 1 design 2 design 3 ... the director discovers she only has 20 blue shirts. what is the greatest arrangement design the director can use before she runs out of blue shirts? enter the answer in the space provided. use numbers instead of words.

Explanation:

First, we analyze the designs. Let's assume the pattern of blue shirts per row. From the given designs (Design 1: maybe 0 blue, Design 2: 2 blue, Design 3: 2 blue? Wait, maybe the pattern is triangular numbers with blue shirts. Wait, the key is that the director has 20 blue shirts. Let's find the largest n such that the sum of blue shirts in each row (maybe following a pattern) is ≤20. Wait, maybe the arrangement is a triangle where each row (after the first) has a certain number of blue shirts. Wait, looking at Design 2: 2 blue, Design 3: 2 blue? No, maybe the pattern is that the number of blue shirts per row follows an arithmetic sequence or triangular numbers. Wait, maybe the formula for the number of blue shirts in a design with k rows (after the first) is the sum of an arithmetic series. Wait, let's re - examine.

Wait, the front row has red (1 red). Then the next rows: Design 2: 2 blue, Design 3: 2 blue? No, maybe the number of blue shirts in each non - front row is increasing. Wait, maybe the pattern is that for a design with n rows (excluding the front red row), the number of blue shirts is the sum from i = 1 to n of i? No, that's triangular numbers. Wait, let's check the total blue shirts. Let's assume that the number of blue shirts in a design with k "blue - row groups" is given by the sum of an arithmetic sequence. Wait, maybe the correct approach is to find the largest n such that the sum of the first n positive integers (triangular number) is ≤20? Wait, no. Wait, maybe the blue shirts are arranged in rows where each row (after the first) has one more blue shirt than the previous. Wait, let's calculate triangular numbers:

Triangular number formula: $T_n=\frac{n(n + 1)}{2}$

We need to find the largest n such that $T_n\leq20$

Step 1: Test n = 5

$T_5=\frac{5\times(5 + 1)}{2}=\frac{5\times6}{2}=15$

Step 2: Test n = 6

$T_6=\frac{6\times(6 + 1)}{2}=\frac{6\times7}{2}=21$

Since 21>20 and 15≤20, but wait, maybe the pattern is different. Wait, maybe the front row is red (1 red), and then the blue rows: Design 2 has 2 blue, Design 3 has 2 blue? No, maybe the number of blue shirts per row is 2, 2, 3? No, the problem says "the greatest arrangement design" with 20 blue shirts. Wait, maybe the number of blue shirts in each row (after the first) is 2, 4, 6... No, that's even numbers. Wait, maybe the correct pattern is that the number of blue shirts in a design with k rows (of blue) is 2 + 4+...+2k? No, that's 2(1 + 2+...+k)=k(k + 1). Let's check:

For k = 4: 4×5 = 20. Oh! That's exactly 20.

Wait, let's see: if the number of blue shirts in each row (of blue) is 2, 4, 6, 8... Wait, no, 2+4 + 6+8=20. Wait, 2+4=6, 2+4+6 = 12, 2+4+6+8 = 20. So that's 4 rows of blue shirts (with 2,4,6,8 blue shirts respectively) and 1 front row of red. But maybe the design number is related to the number of blue - row groups. Wait, Design 2: maybe 2 blue shirts (1 row of blue), Design 3: 2 + 4=6 blue shirts (2 rows of blue)? No, the problem is to find the greatest design number where the total blue shirts ≤20.

Wait, let's re - express. Let's assume that the number of blue shirts in Design n is the sum of the first (n - 1) even numbers. The sum of the first m even numbers is m(m + 1). Wait, for m = 4: 4×5=20. So if Design n has (n - 1) rows of blue shirts, and the sum of blue shirts is (n - 1)n. We need (n - 1)n≤20.

Solve (n - 1)n≤20:

n² - n - 20≤0

The roots of n² - n - 20 = 0 are $n=\frac{1\pm\sqrt{1 + 80}}{2}=\frac{1\pm9}{2}$. Positive root is $\frac{1 + 9}{2}=5$. So when n = 5, (5 - 1)×5=20, which is equal to 20. When n = 6, (6 - 1)×6 = 30>20. So the greatest design number is 5? Wait, no, maybe the pattern is different. Wait, looking at the designs:

Design 1: Maybe 0 blue (front row red only)

Design 2: 2 blue (1 row of blue, 2 shirts)

Design 3: 2+4 = 6 blue (2 rows of blue, 2 and 4 shirts)

Design 4: 2+4+6 = 12 blue (3 rows of blue, 2,4,6 shirts)

Design 5: 2+4+6+8 = 20 blue (4 rows of blue, 2,4,6,8 shirts)

Yes! So the sum of blue shirts for Design n (where n≥2) is the sum of the first (n - 1) even numbers. The sum of the first m even numbers is m(m + 1). Wait, for m = 4 (n - 1=4⇒n = 5), the sum is 4×5 = 20. So the greatest design number is 5.

Answer:

5