kittykat

To determine which integers \( n \) in the range \( 70 \leq n \leq 90 \) can be expressed in the form

\[
n = ab + 2a + 3b,
\]

we can rearrange this equation:

\[
n = ab + 2a + 3b = ab + 2a + 3b = a(b + 2) + 3b.
\]

Thus, we can factor the expression:

\[
n = a(b + 2) + 3b = (b + 2)a + 3b.
\]

Rearranging further, we can consider the expression more carefully:

\[
n = ab + 2a + 3b = a(b + 2) + 3b.
\]

To find how \( n \) varies based on \( a \) and \( b \), we can think of expressions:

\[
n = ab + 2a + 3b = a(b + 2) + 3b.
\]

To explore values of \( n \), we can set specific values for \( b \) and investigate how \( n \) changes. We start by expressing \( a \) in terms of \( n \) for different values of \( b \):

1. Rearranging gives us:

\[
n - 3b = a(b + 2).
\]

This implies

\[
a = \frac{n - 3b}{b + 2}.
\]

For \( a \) to be a positive integer, \( n - 3b \) must be divisible by \( b + 2 \) and \( n - 3b > 0 \). This leads to key constraints:

- \( n > 3b \) (or equivalently, \( b < \frac{n}{3} \)),

- \( n - 3b \) should be divisible by \( b + 2 \).

Now, let’s analyze integers within the specified range \( 70 \leq n \leq 90 \).

**For \( b = 1 \):**

\[
n = a(1 + 2) + 3 \cdot 1 = 3a + 3 \Rightarrow n - 3 = 3a \Rightarrow n = 3a + 3 \Rightarrow n - 3 \equiv 0 \, (\text{mod } 3).
\]

Thus, \( n \equiv 0 \, (\text{mod } 3) \) which gives us possible \( n \): \( 72, 75, 78, 81, 84, 87, 90 \).

**For \( b = 2 \):**

\[
n = a(2 + 2) + 3 \cdot 2 = 4a + 6 \Rightarrow n - 6 = 4a \Rightarrow n = 4a + 6 \Rightarrow n - 6 \equiv 0 \, (\text{mod } 4).
\]

Thus, \( n \equiv 2 \, (\text{mod } 4) \) which gives us \( n \): \( 70, 74, 78, 82, 86, 90 \).

**For \( b = 3 \):**

\[
n = a(3 + 2) + 3 \cdot 3 = 5a + 9 \Rightarrow n - 9 = 5a \Rightarrow n = 5a + 9 \Rightarrow n - 9 \equiv 0 \, (\text{mod } 5).
\]

Thus, \( n \equiv 4 \, (\text{mod } 5) \) which gives us \( n \): \( 74, 79, 84, 89 \).

**For \( b = 4 \):**

\[
n = a(4 + 2) + 3 \cdot 4 = 6a + 12 \Rightarrow n - 12 = 6a \Rightarrow n = 6a + 12 \Rightarrow n - 12 \equiv 0 \, (\text{mod } 6).
\]

Thus, \( n \equiv 0 \, (\text{mod } 6) \) which gives us \( n \): \( 72, 78, 84, 90 \).

Continuing this process for \( b = 5 \) and \( b = 6 \) eventually provides specific integers. We can compile our results and find the integers \( n \):

By synthesizing:

- All candidate results from \( b = 1, 2, 3, 4, 5, 6 \):

The integers derived include:

- \( n = 70, 72, 74, 75, 78, 79, 81, 82, 84, 86, 87, 89, 90 \).

Finally, we count unique integers:

\[
\{70, 72, 74, 75, 78, 79, 81, 82, 84, 86, 87, 89, 90\}
\]

Counting these gives:

\[
\text{Total count} = 13.
\]

Thus, the answer is

\[
\boxed{13}.
\]

We want to find the exponential form of the complex number:

\[
e^{\frac{17\pi i}{60}} + e^{\frac{27\pi i}{60}} + e^{\frac{37\pi i}{60}} + e^{\frac{47\pi i}{60}} + e^{\frac{57\pi i}{60}}
\]

### Step 1: Recognize the terms as roots of unity

The terms \( e^{\frac{17\pi i}{60}}, e^{\frac{27\pi i}{60}}, e^{\frac{37\pi i}{60}}, e^{\frac{47\pi i}{60}}, e^{\frac{57\pi i}{60}} \) are all complex numbers in exponential form. Notice that these exponents can be expressed as:

\[
\frac{17\pi i}{60}, \frac{27\pi i}{60}, \frac{37\pi i}{60}, \frac{47\pi i}{60}, \frac{57\pi i}{60}
\]

These correspond to angles \( \theta = \frac{17\pi}{60}, \frac{27\pi}{60}, \frac{37\pi}{60}, \frac{47\pi}{60}, \frac{57\pi}{60} \).

Since \( e^{i\theta} \) represents a point on the unit circle in the complex plane, each of these terms can be considered as specific roots of unity, although not all are primitive roots.

### Step 2: Consider the sum of the angles

The sum can be expressed as:

\[
S = e^{\frac{17\pi i}{60}} + e^{\frac{27\pi i}{60}} + e^{\frac{37\pi i}{60}} + e^{\frac{47\pi i}{60}} + e^{\frac{57\pi i}{60}}
\]

These angles are evenly spaced on the unit circle by an angle increment of \( \frac{10\pi}{60} = \frac{\pi}{6} \).

### Step 3: Utilize symmetry

These angles correspond to the roots of the equation \( x^5 - 1 = 0 \) rotated by a small angle \( \frac{17\pi}{60} \). The angles \( \frac{17\pi}{60}, \frac{27\pi}{60}, \frac{37\pi}{60}, \frac{47\pi}{60}, \frac{57\pi}{60} \) map to the five vertices of a regular pentagon inscribed in the unit circle but rotated slightly.

For a regular \( n \)-gon (in this case, \( n = 5 \)), the sum of vectors corresponding to the vertices is zero if the vectors are evenly distributed around the circle (because they symmetrically cancel each other out). However, here, the vertices are slightly rotated, but they are still symmetrically spaced around the circle.

### Step 4: Apply properties of roots of unity

Given the symmetry and spacing:

\[
S = e^{i\frac{\theta_0}{60}} \cdot \left(1 + e^{i\frac{2\pi}{5}} + e^{i\frac{4\pi}{5}} + e^{i\frac{6\pi}{5}} + e^{i\frac{8\pi}{5}}\right)
\]

Where \( \theta_0 \) is \( \frac{17\pi}{60} \), and the roots sum to zero because they represent the vertices of a pentagon. Thus, the sum

:

\[
S = 0
\]

### Conclusion:

The exponential form of the given complex number is:

\[
\boxed{0}
\]

Given that triangle \(ABC\) has altitudes \(AD = 14\), \(BE = 16\), and \(CF = h\), where \(h\) is a positive integer, we need to find the largest possible value of \(h\).

### Step 1: Use the area formula with altitudes

The area of triangle \(ABC\) can be expressed in terms of the altitudes and the corresponding sides:

\[
\text{Area} = \frac{1}{2} \times a \times h_a = \frac{1}{2} \times b \times h_b = \frac{1}{2} \times c \times h_c
\]

where \(h_a = AD\), \(h_b = BE\), and \(h_c = CF = h\), and \(a\), \(b\), and \(c\) are the lengths of the sides opposite the vertices \(A\), \(B\), and \(C\), respectively.

### Step 2: Express the sides in terms of the area and altitudes

Let the area of triangle \(ABC\) be denoted as \(K\). Then:

\[
K = \frac{1}{2} \times a \times 14 = 7a
\]

\[
K = \frac{1}{2} \times b \times 16 = 8b
\]

\[
K = \frac{1}{2} \times c \times h = \frac{1}{2}ch
\]

### Step 3: Equate the expressions for the area \(K\)

From the first two equations:

\[
7a = 8b \quad \Rightarrow \quad a = \frac{8b}{7}
\]

Using the equation for \(K\) involving \(h\):

\[
7a = \frac{1}{2} ch
\]

Substitute \(a = \frac{8b}{7}\) into the equation:

\[
7\left(\frac{8b}{7}\right) = \frac{1}{2}ch
\]

Simplify:

\[
8b = \frac{1}{2}ch
\]

Multiply both sides by 2:

\[
16b = ch
\]

Thus:

\[
h = \frac{16b}{c}
\]

### Step 4: Maximize \(h\)

We aim to maximize \(h = \frac{16b}{c}\), where \(b\) and \(c\) are positive integers.

Recall that \(K = 7a = 8b = \frac{1}{2}ch\). We want to maximize \(h\), so:

\[
c = \frac{16b}{h}
\]

To maximize \(h\), minimize \(c\). Since \(b = 7\), \(a = 8\), choose \(b = 7\) and \(c = 1\) which gives:

\[
h = \frac{16 \times 7}{1} = 112
\]

However, since \(b = \gcd(7,8)\), choose \(b = 7\) for simplicity. So:

\[
K = 112, h = 1
]

Thus:

\[
h \leq 112
\]

### Final Answer:

Since \(h = 112\), the maximum value for the given maximum number is \( \boxed{112} \).

To determine the number of points of the form \((x,y)\), where both coordinates are positive integers, that lie below the graph of the hyperbola \(xy = 16\), we need to find the integer pairs \((x, y)\) such that \(xy < 16\).

### Step-by-Step Solution:

1. **Identify the Range for \(x\)**:

   - For each \(x\), we need \(y\) to be an integer such that \(xy < 16\).

   - Since \(x\) must be a positive integer, consider possible values of \(x\) starting from 1 up to the point where \(x \cdot 1 = 16\), so \(x\) ranges from 1 to 15.

2. **Count \(y\) Values for Each \(x\)**:

   - For each \(x\), find the largest integer \(y\) such that \(y < \frac{16}{x}\).

   Here’s how this works for each \(x\) from 1 to 15:

   - \(x = 1\): \(xy < 16 \implies y < \frac{16}{1} = 16\). So, \(y\) can be 1 to 15 (15 values).

   - \(x = 2\): \(xy < 16 \implies y < \frac{16}{2} = 8\). So, \(y\) can be 1 to 7 (7 values).

   - \(x = 3\): \(xy < 16 \implies y < \frac{16}{3} \approx 5.33\). So, \(y\) can be 1 to 5 (5 values).

   - \(x = 4\): \(xy < 16 \implies y < \frac{16}{4} = 4\). So, \(y\) can be 1 to 3 (3 values).

   - \(x = 5\): \(xy < 16 \implies y < \frac{16}{5} = 3.2\). So, \(y\) can be 1 to 3 (3 values).

   - \(x = 6\): \(xy < 16 \implies y < \frac{16}{6} \approx 2.67\). So, \(y\) can be 1 to 2 (2 values).

   - \(x = 7\): \(xy < 16 \implies y < \frac{16}{7} \approx 2.29\). So, \(y\) can be 1 to 2 (2 values).

   - \(x = 8\): \(xy < 16 \implies y < \frac{16}{8} = 2\). So, \(y\) can be 1 (1 value).

   - \(x = 9\) to \(x = 15\): For these values, \(y < \frac{16}{x}\) will always be less than 2, so \(y\) can only be 1 (1 value each).

3. **Summarize the Counts**:

   \[
   \begin{aligned}

   &15 \text{ values for } x = 1, \\

   &7 \text{ values for } x = 2, \\

   &5 \text{ values for } x = 3, \\

   &3 \text{ values for } x = 4, \\

   &3 \text{ values for } x = 5, \\

   &2 \text{ values for } x = 6, \\

   &2 \text{ values for } x = 7, \\

   &1 \text{ value for } x = 8, \\

   &1 \text{ value each for } x = 9 \text{ to } 15.

   \end{aligned}

   \]

   Adding these up:

   \[
   15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 44
   \]

Therefore, the number of points \((x, y)\), where both coordinates are positive integers, that lie below the graph of the hyperbola \(xy = 16\) is:
\[
\boxed{44}
\]

Five CDs will cost 55 dollars.

We can find cos(C) using the Law of Cosines. Here's how:

Law of Cosines:

The Law of Cosines relates the sides and angles of a triangle. For triangle ABC, it states:

c^2 = a^2 + b^2 - 2ab * cos(C)

where:

c is the side opposite angle C (in this case, side AB)

a is the side adjacent to angle A (in this case, side BC)

b is the side adjacent to angle B (in this case, side AC)

Given Information:

We are given:

cos(A) = sqrt(7/10)

cos(B) = sqrt(3/10)

We need to find cos(C).

Sides are Unknown:

We don't have information about the side lengths (a, b, or c). However, we can express them using cos(A) and cos(B) for further calculations.

Expressing Sides:

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can find sin^2(A) and sin^2(B):

sin^2(A) = 1 - cos^2(A) = 1 - (sqrt(7/10))^2 = 3/10

sin^2(B) = 1 - cos^2(B) = 1 - (sqrt(3/10))^2 = 7/10

Relating Sides to Cosines:

Since we are looking for cos(C), let's focus on side c (AB). Using cosine definitions in right triangles:

cos(A) = (adjacent side) / (hypotenuse) = a / c

cos(B) = (adjacent side) / (hypotenuse) = b / c

Therefore:

a = c * cos(A) = c * sqrt(7/10)

b = c * cos(B) = c * sqrt(3/10)

Applying the Law of Cosines:

Substitute the expressions for a and b in the Law of Cosines equation:

c^2 = (c * sqrt(7/10))^2 + (c * sqrt(3/10))^2 - 2 * c * sqrt(7/10) * c * sqrt(3/10) * cos(C)

Simplifying the equation:

c^2 = (7/10)c^2 + (3/10)c^2 - (6/10)c^2 * cos(C)

Combining like terms:

c^2 = (7 + 3 - 6) / 10 * c^2 - (6/10)c^2 * cos(C)

4c^2 / 10 = -(6/10)c^2 * cos(C)

Solving for cos(C):

Isolate cos(C):

cos(C) = -(4c^2 / 10) / -(6/10)c^2

cos(C) = 4/6 (Since both c^2 terms are negative, the negative signs cancel out)

cos(C) = 2/3

Therefore, cos(C) = 2/3.

To solve this problem, let's first find the total number of possible pairs of cards that Professor Grok can draw. Since there are 4 colors and 7 numbers, there are \(4 \times 7 = 28\) cards in total.

Now, let's find the number of pairs where the first card drawn is labeled with an even number and the second card drawn is labeled with a multiple of 3.

1. There are 7 even numbers among 1 to 7: 2, 4, 6.

2. There are 7 multiples of 3 among 1 to 7: 3, 6.

To find the number of pairs with the desired properties:

- For the first card, there are 7 even numbers, and for the second card, there are 7 multiples of 3.

- So, the total number of pairs with the desired properties is \(7 \times 7 = 49\).

Now, to find the probability, divide the number of pairs with the desired properties by the total number of possible pairs:

\[\text{Probability} = \frac{\text{Number of pairs with desired properties}}{\text{Total number of possible pairs}} = \frac{49}{28 \times 27}\]

However, since Professor Grok is drawing without replacement, after the first card is drawn, there are only 27 cards left. So, the denominator should be \(28 \times 27\). 

Thus, the probability that the first card Grok draws is labeled with an even number, and the second card Grok draws is labeled with a multiple of 3 is:

\[\frac{49}{28 \times 27}  = \frac{7}{108}.\]

Problem: Let a_1, a_2, a_3 be real numbers such that

|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_1| = 100.
What is the largest possible value of |a_1 - a_2|?

Answer: 10

Here's how to find the radius of the circle:

Area of a Single White Sector:

We are given that the area of one white sector is 13π/54 square inches.

Proportion of White Sectors:

Since there are 21 sectors total and 3 are white, the white sectors represent 3/21 of the entire circle's area.

Full Circle Area:

To find the total area of the circle, we can divide the area of one white sector by its proportion (3/21) of the whole circle:

Total circle area = (Area of one white sector) / (Proportion of white sectors)

Total circle area = (13π/54 in²) / (3/21)

Circle Area Formula:

We know the area of a circle is calculated by πr², where r is the radius.

Equating Areas:

Since the total area we found represents the entire circle, we can equate it to the circle area formula:

πr² = (13π/54 in²) / (3/21)

Simplifying and Solving:

Cancel out common factors of π and simplify:

r² = (13 / 54) * (21 / 3) in²

r² = 13 in²

Take the square root of both sides (remembering that squaring can introduce extraneous solutions, so we'll check for those later).

r = √13 in ≈ 3.61 in (rounded to nearest hundredth)