mercurym999

Five positive integers are arranged in a circle, with residues 0, 1, or 2 modulo 3. No two adjacent numbers (a_i + a_(i+1)) or three adjacent numbers (a_i + a_(i+1) + a_(i+2)) have a sum divisible by 3, indices modulo 5. We need the number of a_i ≡ 0 (mod 3), with options 0, 1, 2, or 3.

Exactly 2 numbers are divisible by 3. The sequence 0, 1, 0, 1, 1:

Two adjacent: 0+1=1, 1+0=1, 0+1=1, 1+1=2, 1+0=1. All ≢ 0 (mod 3).

Three adjacent: 0+1+0=1, 1+0+1=2, 0+1+1=2, 1+1+0=2, 1+0+1=2. All ≢ 0 (mod 3).

Other cases fail:

Zero 0’s: E.g., 1, 1, 1, 1, 1: 1+1+1=3 ≡ 0. Or 1, 2, 1, 2, 1: 1+2=3 ≡ 0.

One 0: E.g., 0, 1, 1, 1, 1: 1+1+1=3 ≡ 0.

Three 0’s: E.g., 0, 1, 0, 1, 0: 0+0=0 (mod 3).

Thus, 2 is the only solution.

Answer: (C) 2

To solve (ax^2 - 24x + b)/(x - 1) = x^2 + x, which holds for two real x-values summing to 12, and find a - b:

Multiply by x - 1:

ax^2 - 24x + b = (x^2 + x)(x - 1)

Expand:

(x^2 + x)(x - 1) = x^3 - x

So:

ax^2 - 24x + b = x^3 - x

Move terms:

x^3 - x - (ax^2 - 24x + b) = x^3 - ax^2 + 23x - b = 0

Cubic:

x^3 - ax^2 + 23x - b = 0

Assume roots p, p, q (p double root). Vieta's gives:

2p + q = a

Distinct roots p, q sum to 12:

p + q = 12

So, q = 12 - p. Cubic as (x - p)^2 (x - q):

(x - p)^2 (x - q) = x^3 - (2p + q)x^2 + (p^2 + 2pq)x - p^2 q

Compare:

x^2: -(2p + q) = -a, so a = 2p + q

x: p^2 + 2pq = 23

Constant: -p^2 q = -b, so b = p^2 q

Solve:

p^2 + 2pq = 23

With q = 12 - p:

p^2 + 2p(12 - p) = -p^2 + 24p = 23

p^2 - 24p + 23 = 0

p = (24 ± sqrt(576 - 92))/2 = (24 ± 22)/2

p = 23 or 1

Take p = 1:

q = 12 - 1 = 11

a = 2 * 1 + 11 = 13

b = 1^2 * 11 = 11

a - b = 13 - 11 = 2

Cubic x^3 - 13x^2 + 23x - 11 = 0 has roots x = 1 (double), x = 11, summing to 12.

Answer: a - b = 2

We can use the combination formula: C(n, k) = n! / (k! * (n-k)!), where n is the total number of people (8) and k is the number to choose (2). So, we need C(8, 2) = 8! / (2! * 6!).

Calculate the factorials:

8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

6! = 6 * 5 * 4 * 3 * 2 * 1

2! = 2 * 1 = 2

Plugging in: C(8, 2) = (8 * 7 * 6!) / (2 * 1 * 6!). The 6! cancels out, leaving (8 * 7) / (2 * 1) = 56 / 2 = 28.

Alternatively, the formula simplifies to C(n, k) = [n * (n-1) * ... * (n-k+1)] / k!. For n=8, k=2: C(8, 2) = (8 * 7) / (2 * 1) = 56 / 2 = 28.

mercurym999

For three consecutive tiles with numbers \(a_i, a_{i+1}, a_{i+2}\), we require \(a_i + a_{i+1} + a_{i+2} = 3\).

Since each number is in {-1, 0, 1}, the maximum sum is \(1 + 1 + 1 = 3\), so \(a_i = a_{i+1} = a_{i+2} = 1\).

So, all 7 tiles must have number 1 to satisfy for all consecutive triples.

Each of the 7 tiles, all numbered 1, can be red or blue, giving \(2^7 = 128\) color assignments.

So there are 128 possible arrangements.

mercurym999

For points P₁ = (cos 37°, sin 37°) and P₂ = (cos 127°, sin 127°), the distance is:

√((cos 127° - cos 37°)² + (sin 127° - sin 37°)²).

The angle difference: 127° - 37° = 90°. Simplify:

(cos 127° - cos 37°)² + (sin 127° - sin 37°)².

Expand:

= cos² 127° - 2 cos 127° cos 37° + cos² 37° + sin² 127° - 2 sin 127° sin 37° + sin² 37°.

Since cos² θ + sin² θ = 1:

= (1 + 1) - 2 (cos 127° cos 37° + sin 127° sin 37°).

Use the cosine angle sum identity:

cos 127° cos 37° + sin 127° sin 37° = cos (127° - 37°) = cos 90° = 0.

Thus:

= 2 - 2 · 0 = 2.

Distance:

√2.

hmmm im getting $3 \pi,$ anyway i used wolframalpha for \[\left| z + \frac{1}{z} \right| \le 2.\] and I got this